› ~wylee › 2012_CHL.pdf · Hyponormality and Subnormality of Block Toeplitz Operators Raul E....

Hyponormality and Subnormality of Block Toeplitz Operators

Raul E. Curto, In Sung Hwang and Woo Young Lee

Abstract. In this paper we are concerned with hyponormality and subnormality of block Toeplitzoperators acting on the vector-valued Hardy space H2

Cn of the unit circle.Firstly, we establish a tractable and explicit criterion on the hyponormality of block

Toeplitz operators having bounded type symbols via the triangularization theorem for com-pressions of the shift operator.

Secondly, we consider the gap between hyponormality and subnormality for block Toeplitzoperators. This is closely related to Halmos’s Problem 5: Is every subnormal Toeplitz operatoreither normal or analytic ? We show that if Φ is a matrix-valued rational function whose co-analytic part has a coprime factorization then every hyponormal Toeplitz operator TΦ whosesquare is also hyponormal must be either normal or analytic.

Thirdly, using the subnormal theory of block Toeplitz operators, we give an answer to thefollowing “Toeplitz completion” problem: Find the unspecified Toeplitz entries of the partialblock Toeplitz matrix

A :=

[U∗ ?? U∗

]so that A becomes subnormal, where U is the unilateral shift on H2.

Keywords. Block Toeplitz operators, hyponormal, square-hyponormal, subnormal, boundedtype functions, rational functions, trigonometric polynomials, subnormal completion problem.

1. Introduction

Toeplitz operators, block Toeplitz operators and (block) Toeplitz determinants (i.e., determinantsof sections of (block) Toeplitz operators) arise naturally in several fields of mathematics and ina variety of problems in physics, especially, in quantum mechanics. For example, the spectraltheory of Toeplitz operators plays an important role in the study of solvable models in quantummechanics ([Pr]) and in the study the one-dimensional Heisenberg Hamiltonian of ferromagnetism([DMA]); the theory of block Toeplitz determinants is used in the study of the classical dimermodel ([BE]) and in the study of the vicious walker model ([HI]); the theory of block Toeplitzoperators is also used in the study of Gelfand-Dickey Hierarchies (cf. [Ca]). On the other hand,the theory of hyponormal and subnormal operators is an extensive and highly developed area,which has made important contributions to a number of problems in functional analysis, operatortheory, and mathematical physics (see, for example, [If], [HS], and [Sz] for applications to relatedmathematical physics problems). Thus, it becomes of central significance to describe in detailhypormality and subnormality for Toeplitz operators. This paper focuses primarily on hyponor-mality and subnormality of block Toeplitz operators with rational symbols. For the general theoryof subnormal and hyponormal operators, we refer to [Con] and [MP].

To describe our results, we first need to review a few essential facts about (block) Toeplitzoperators, and for that we will use [BS], [Do1], [Do2], [GGK], [MAR], [Ni], and [Pe]. Let H and K

2010 Mathematics Subject Classification. Primary 47B20, 47B35, 47A13; Secondary 30H10, 47A20, 47A57The work of the first named author was partially supported by NSF Grant DMS-0801168. The work of the second

author was supported by NRF grant funded by the Korea government(MEST)(2011-0003273). The work of thethird author was supported by Basic Science Research Program through the NRF grant funded by the Koreagovernment(MEST)(2011-0001250).

2 Raul E. Curto, In Sung Hwang and Woo Young Lee

be complex Hilbert spaces, let B(H,K) be the set of bounded linear operators from H to K, andwrite B(H) := B(H,H). For A,B ∈ B(H), we let [A,B] := AB − BA. An operator T ∈ B(H)is said to be normal if [T ∗, T ] = 0, hyponormal if [T ∗, T ] ≥ 0, and subnormal if T has a normalextension, i.e., T = N |H, where N is a normal operator on some Hilbert space K ⊇ H such thatH isinvariant for N . For an operator T ∈ B(H), we write kerT and ranT for the kernel and the range ofT , respectively. For a setM, clM andM⊥ denote the closure and the orthogonal complement ofM, respectively. Also, let T = R/2πZ be the unit circle. Recall that the Hilbert space L2 ≡ L2(T)has a canonical orthonormal basis given by the trigonometric functions en(z) = zn, for all n ∈ Z,and that the Hardy space H2 ≡ H2(T) is the closed linear span of en : n = 0, 1, . . .. An elementf ∈ L2 is said to be analytic if f ∈ H2. Let H∞ ≡ H∞(T) := L∞ ∩ H2, i.e., H∞ is the set ofbounded analytic functions on the open unit disk D.

Given a bounded measurable function ϕ ∈ L∞, the Toeplitz operator Tϕ and the Hankeloperator Hϕ with symbol ϕ on H2 are defined by

Tϕg := P (ϕg) and Hϕg := JP⊥(ϕg) (g ∈ H2), (1.1)

where P and P⊥ denote the orthogonal projections that map from L2 onto H2 and (H2)⊥, re-spectively, and J denotes the unitary operator from L2 onto L2 defined by J(f)(z) = zf(z) forf ∈ L2.

To study hyponormality (resp. normality and subnormality) of the Toeplitz operator Tϕ withsymbol ϕ we may, without loss of generality, assume that ϕ(0) = 0; this is because hyponormality(resp. normality and subnormality) is invariant under translations by scalars. Normal Toeplitzoperators were characterized by a property of their symbols in the early 1960’s by A. Brown andP.R. Halmos [BH] and the exact nature of the relationship between the symbol ϕ ∈ L∞ and thehyponormality of Tϕ was understood via Cowen’s Theorem [Co4] in 1988.

Cowen’s Theorem. ([Co4], [NT]) For each ϕ ∈ L∞, let

E(ϕ) ≡ k ∈ H∞ : ||k||∞ ≤ 1 and ϕ− kϕ ∈ H∞.

Then a Toeplitz operator Tϕ is hyponormal if and only if E(ϕ) is nonempty.

This elegant and useful theorem has been used in the works [CuL1], [CuL2], [FL], [Gu1],[Gu2], [GS], [HKL1], [HKL2], [HL1], [HL2], [HL3], [Le], [NT] and [Zhu], which have been devotedto the study of hyponormality for Toeplitz operators on H2. Particular attention has been paid toToeplitz operators with polynomial symbols or rational symbols [HL2], [HL3]. However, the case ofarbitrary symbol ϕ, though solved in principle by Cowen’s theorem, is in practice very complicated.Indeed, it may not even be possible to find tractable necessary and sufficient condition for thehyponormality of Tϕ in terms of the Fourier coefficients of the symbol ϕ unless certain assumptionsare made about ϕ. To date, tractable criteria for the cases of trigonometric polynomial symbolsand rational symbols were derived from a Caratheodory-Schur interpolation problem ([Zhu]) and atangential Hermite-Fejer interpolation problem ([Gu1]) or the classical Hermite-Fejer interpolationproblem ([HL3]), respectively.

Recall that a function ϕ ∈ L∞ is said to be of bounded type (or in the Nevanlinna class) ifthere are analytic functions ψ1, ψ2 ∈ H∞(D) such that ϕ = ψ1/ψ2 almost everywhere on T. Todate, no tractable criterion to determine the hyponormality of Tϕ when the symbol ϕ is of boundedtype has been found.

We now introduce the notion of block Toeplitz operators. Let Mn×r denote the set of alln × r complex matrices and write Mn := Mn×n. For X a Hilbert space, let L2

X ≡ L2X (T) be the

Hilbert space of X -valued norm square-integrable measurable functions on T and let H2X ≡ H2

X (T)be the corresponding Hardy space. We observe that L2

Cn = L2 ⊗ Cn and H2Cn = H2 ⊗ Cn. If Φ

is a matrix-valued function in L∞Mn≡ L∞Mn

(T) (= L∞ ⊗Mn) then TΦ : H2Cn → H2

Cn denotes theblock Toeplitz operator with symbol Φ defined by

TΦf := Pn(Φf) for f ∈ H2Cn ,

Block Toeplitz Operators 3

where Pn is the orthogonal projection of L2Cn onto H2

Cn . A block Hankel operator with symbolΦ ∈ L∞Mn

is the operator HΦ : H2Cn → H2

Cn defined by

HΦf := JnP⊥n (Φf) for f ∈ H2

Cn ,

where Jn denotes the unitary operator from L2Cn onto L2

Cn given by Jn(f)(z) := zInf(z) forf ∈ L2

Cn , with In the n× n identity matrix. If we set H2Cn = H2 ⊕ · · · ⊕H2 then we see that

TΦ =

Tϕ11 . . . Tϕ1n

...Tϕn1 . . . Tϕnn

and HΦ =

Hϕ11 . . . Hϕ1n

...Hϕn1 . . . Hϕnn

,where

Φ =

ϕ11 . . . ϕ1n

...ϕn1 . . . ϕnn

∈ L∞Mn.

For Φ ∈ L∞Mn×m, write

Φ(z) := Φ∗(z).

A matrix-valued function Θ ∈ H∞Mn×m(= H∞ ⊗ Mn×m) is called inner if Θ∗Θ = Im almost

everywhere on T. The following basic relations can be easily derived:

T ∗Φ = TΦ∗ , H∗Φ = HΦ (Φ ∈ L∞Mn); (1.2)

TΦΨ − TΦTΨ = H∗Φ∗HΨ (Φ,Ψ ∈ L∞Mn); (1.3)

HΦTΨ = HΦΨ, HΨΦ = T ∗ΨHΦ (Φ ∈ L∞Mn

,Ψ ∈ H∞Mn); (1.4)

H∗ΦHΦ −H∗ΘΦHΘΦ = H∗ΦHΘ∗H∗Θ∗HΦ (Θ ∈ H∞Mn

inner, Φ ∈ L∞Mn). (1.5)

For a matrix-valued function Φ ≡ [ϕij ] ∈ L∞Mn, we say that Φ is of bounded type if each entry

ϕij is of bounded type, and we say that Φ is rational if each entry ϕij is a rational function. Amatrix-valued trigonometric polynomial Φ ∈ L∞Mn

is of the form

Φ(z) =

N∑j=−m

Ajzj (Aj ∈Mn),

where AN and A−m are called the outer coefficients of Φ.We recall that for matrix-valued functionsA :=

∑∞j=−∞Ajz

j ∈ L2Mn

andB :=∑∞j=−∞Bjz

j ∈L2Mn

, we define the inner product of A and B by

(A,B) :=

∫T

tr (B∗A) dµ =

∞∑j=−∞

tr (B∗jAj) ,

where tr (·) denotes the trace of a matrix and define ||A||2 := (A,A)12 . We also define, for A ∈ L∞Mn

,

||A||∞ := ess supt∈T||A(t)|| (|| · || denotes the spectral norm of a matrix).

Finally, the shift operator S on H2Cn is defined by

S := TzIn .

The following fundamental result will be useful in the sequel.

The Beurling-Lax-Halmos Theorem. A nonzero subspace M of H2Cn is invariant for the shift opera-

tor S on H2Cn if and only if M = ΘH2

Cm , where Θ is an inner matrix function in H∞Mn×m(m ≤ n).

Furthermore, Θ is unique up to a unitary constant right factor; that is, if M = ∆H2Cr (for ∆ an

inner function in H∞Mn×r), then m = r and Θ = ∆W , where W is a (constant in z) unitary matrix

mapping Cm onto Cm.


As customarily done, we say that two matrix-valued functions A and B are equal if theyare equal up to a unitary constant right factor. Observe by (1.4) that for Φ ∈ L∞Mn

, HΦS =HΦTzIn = HΦ·zIn = HzIn·Φ = T ∗zInHΦ, which implies that the kernel of a block Hankel operator

HΦ is an invariant subspace of the shift operator on H2Cn . Thus, if kerHΦ 6= 0, then by the

Beurling-Lax-Halmos Theorem,

kerHΦ = ΘH2Cm

for some inner matrix function Θ. We note that Θ need not be a square matrix.On the other hand, recently C. Gu, J. Hendricks and D. Rutherford [GHR] considered the

hyponormality of block Toeplitz operators and characterized it in terms of their symbols. Inparticular they showed that if TΦ is a hyponormal block Toeplitz operator on H2

Cn , then its symbolΦ is normal, i.e., Φ∗Φ = ΦΦ∗. Their characterization for hyponormality of block Toeplitz operatorsresembles Cowen’s Theorem except for an additional condition – the normality condition of thesymbol.

Hyponormality of Block Toeplitz Operators ([GHR]) For each Φ ∈ L∞Mn, let

E(Φ) :=K ∈ H∞Mn

: ||K||∞ ≤ 1 and Φ−KΦ∗ ∈ H∞Mn

.

Then TΦ is hyponormal if and only if Φ is normal and E(Φ) is nonempty.

The hyponormality of the Toeplitz operator TΦ with arbitrary matrix-valued symbol Φ,though solved in principle by Cowen’s Theorem [Co4] and the criterion due to Gu, Hendricks andRutherford [GHR], is in practice very complicated. Until now, explicit criteria for the hyponor-mality of block Toeplitz operators TΦ with matrix-valued trigonometric polynomials or rationalfunctions Φ were established via interpolation problems ([GHR], [HL4], [HL5]).

In Section 3, we obtain a tractable criterion for the hyponormality of block Toeplitz operatorswith bounded type symbols. To do this we employ a continuous analogue of the elementary theoremof Schur on triangularization of finite matrices: If T is a finite matrix then it can be representedas T = D + N , where D is a diagonal matrix and N is a nilpotent matrix. The continuousanalogue is the so-called triangularization theorem for compressions of the shift operator: in thiscase, D and N are replaced by a certain (normal) multiplication operator and a Volterra operatorof Hilbert-Schmidt class, respectively.

Section 4 deals with the gap between hyponormality and subnormality of block Toeplitzoperators. The Bram-Halmos criterion for subnormality ([Br], [Con]) states that an operator T ∈B(H) is subnormal if and only if

∑i,j(T

ixj , Tjxi) ≥ 0 for all finite collections x0, x1, · · · , xk ∈ H.

It is easy to see that this is equivalent to the following positivity test:[T ∗, T ] [T ∗2, T ] . . . [T ∗k, T ][T ∗, T 2] [T ∗2, T 2] . . . [T ∗k, T 2]

......

. . ....

[T ∗, T k] [T ∗2, T k] . . . [T ∗k, T k]

≥ 0 (all k ≥ 1). (1.6)

The positivity condition (1.6) for k = 1 is equivalent to the hyponormality of T , while subnormalityrequires the validity of (1.6) for all k ∈ Z+. The Bram-Halmos criterion indicates that hyponor-mality is generally far from subnormality. But there are special classes of operators for whichthe positivity of (1.6) for some k and subnormaity are equivalent. For example, it was shown in([CuL1]) that if W√x,(

√a,√b,√c)∧ is the weighted shift whose weight sequence consists of the initial

weight x followed by the weight sequence of the recursively generated subnormal weighted shiftW(√a,√b,√c)∧ with an initial segment of positive weights

√a,√b,√c (cf. [CuF1], [CuF2], [CuF3]),

then Wα is subnormal if and only if the positivity condition (1.6) is satisfied with k = 2. On theother hand, in [Hal3, Problem 209], it was shown that there exists a hyponormal operator whosesquare is not hyponormal, e.g., U∗ + 2U (U is the unilateral shift on `2), which is a trigonometricToeplitz operator, i.e., U∗+ 2U ≡ Tz+2z. This example addresses the gap between hyponormality


and subnormality for Toeplitz operators. This matter is closely related to Halmos’s Problem 5[Hal1], [Hal2]: Is every subnormal Toeplitz operator either normal or analytic ?

In [CuL1], as a partial answer, it was shown that every hyponormal Toeplitz operator Tϕwith trigonometric polynomial symbol ϕ whose square is hyponormal must be either normal oranalytic. In [Gu3], C. Gu showed that this result still holds for Toeplitz operators Tϕ with rationalsymbol ϕ (more generally, in the cases where ϕ is of bounded type). In Section 4 we prove thefollowing theorem: If Φ is a matrix-valued rational function whose co-analytic part has a coprimefactorization then every hyponormal Toeplitz operator TΦ whose square is hyponormal must beeither normal or analytic. This result generalizes the results in [CuL1] and [Gu3].

In Section 5, we consider a completion problem involving Toeplitz operators. Given a partiallyspecified operator matrix, the problem of finding suitable operators to complete the given partialoperator matrix so that the resulting matrix satisfies certain given properties is called a completionproblem. The dilation problem is a special case of the completion problem: in other words, a dilationof T is a completion of the partial operator matrix

[T ?? ?

]. In recent years, operator theorists have

been interested in the subnormal completion problem for[U∗ ?? U∗

],

where U is the unilateral shift on H2. If the unspecified entries ? are Toeplitz operators this is calledthe Toeplitz subnormal completion problem. Thus this problem is related to the subnormality ofblock Toeplitz operators. In Section 5, we solve this Toeplitz subnormal completion problem.

Finally, in Section 6 we list some open problems.

Acknowledgment. The authors are indebted to the referee for suggestions that improved thepresentation.

2. Basic Theory and Preliminaries

We first recall [Ab, Lemma 3] that if ϕ ∈ L∞ then

ϕ is of bounded type ⇐⇒ kerHϕ 6= 0 . (2.1)

If ϕ ∈ L∞, we write

ϕ+ ≡ Pϕ ∈ H2 and ϕ− ≡ P⊥ϕ ∈ zH2.

Assume now that both ϕ and ϕ are of bounded type. Then from Beurling’s Theorem, kerHϕ− =θ0H

2 and kerHϕ+= θ+H

2 for some inner functions θ0, θ+. We thus have b := ϕ−θ0 ∈ H2, andhence we can write

ϕ− = θ0b and similarly ϕ+ = θ+a for some a ∈ H2. (2.2)

In particular, if Tϕ is hyponormal then since

[T ∗ϕ, Tϕ] = H∗ϕHϕ −H∗ϕHϕ = H∗ϕ+Hϕ+

−H∗ϕ−Hϕ− , (2.3)

it follows that ||Hϕ+f || ≥ ||Hϕ−f || for all f ∈ H2, and hence

θ+H2 = kerHϕ+

⊆ kerHϕ− = θ0H2,

which implies that θ0 divides θ+, i.e., θ+ = θ0θ1 for some inner function θ1. We write, for an innerfunction θ,

H(θ) := H2 θH2.

Note that if f = θa ∈ L2, then f ∈ H2 if and only if a ∈ H(zθ); in particular, if f(0) = 0 thena ∈ H(θ). Thus, if ϕ = ϕ− + ϕ+ ∈ L∞ is such that ϕ and ϕ are of bounded type such thatϕ+(0) = 0 and Tϕ is hyponormal, then we can write

ϕ+ = θ0θ1a and ϕ− = θ0b, where a ∈ H(θ0θ1) and b ∈ H(θ0).


By Kronecker’s Lemma [Ni, p. 183], if f ∈ H∞ then f is a rational function if and only if rankHf <∞, which implies that

f is rational ⇐⇒ f = θb with a finite Blaschke product θ. (2.4)

Also, from the scalar-valued case of (1.4), we can see that if k ∈ E(ϕ) then

[T ∗ϕ, Tϕ] = H∗ϕHϕ −H∗ϕHϕ = H∗ϕHϕ −H∗k ϕHk ϕ = H∗ϕ(1− TkT∗k

)Hϕ. (2.5)

On the other hand, M. Abrahamse [Ab, Lemma 6] showed that if Tϕ is hyponormal, if ϕ /∈ H∞,and if ϕ or ϕ is of bounded type then both ϕ and ϕ are of bounded type. However, by contrast tothe scalar case, Φ∗ may not be of bounded type even though TΦ is hyponormal, Φ /∈ H∞Mn

and Φ isof bounded type. But we have a one-way implication: if TΦ is hyponormal and Φ∗ is of boundedtype then Φ is also of bounded type (see [GHR, Corollary 3.5 and Remark 3.6]). Thus wheneverwe deal with hyponormal Toeplitz operators TΦ with symbols Φ satisfying that both Φ and Φ∗ areof bounded type (e.g., Φ is a matrix-valued rational function), it suffices to assume that only Φ∗ isof bounded type. In spite of this, for convenience, we will assume that Φ and Φ∗ are of boundedtype whenever we deal with bounded type symbols.

For a matrix-valued function Φ ∈ H2Mn×r

, we say that ∆ ∈ H2Mn×m

is a left inner divisor of

Φ if ∆ is an inner matrix function such that Φ = ∆A for some A ∈ H2Mm×r

(m ≤ n). We also say

that two matrix functions Φ ∈ H2Mn×r

and Ψ ∈ H2Mn×m

are left coprime if the only common left

inner divisor of both Φ and Ψ is a unitary constant and that Φ ∈ H2Mn×r

and Ψ ∈ H2Mm×r

are

right coprime if Φ and Ψ are left coprime. Two matrix functions Φ and Ψ in H2Mn

are said to be

coprime if they are both left and right coprime. We note that if Φ ∈ H2Mn

is such that det Φ is

not identically zero then any left inner divisor ∆ of Φ is square, i.e., ∆ ∈ H2Mn

: indeed, if Φ = ∆A

with ∆ ∈ H2Mn×r

(r < n) then for almost all z ∈ T, rank Φ(z) ≤ rank ∆(z) ≤ r < n, so that

det Φ(z) = 0 for almost all z ∈ T. If Φ ∈ H2Mn

is such that det Φ is not identically zero then we say

that ∆ ∈ H2Mn

is a right inner divisor of Φ if ∆ is a left inner divisor of Φ.

On the other hand, we have (in the Introduction) remarked that Θ need not be square inthe equality kerHΦ = ΘH2

Cn , which comes from the Beurling-Lax-Halmos Theorem. But it wasknown [GHR, Theorem 2.2] that for Φ ∈ L∞Mn

, Φ is of bounded type if and only if kerHΦ = ΘH2Cn

for some square inner matrix function Θ.Let Θi ∈ H∞Mn

: i ∈ J be a family of inner matrix functions. Then the greatest common leftinner divisor Θd and the least common left inner multiple Θm of the family Θi ∈ H∞Mn

: i ∈ Jare the inner functions defined by

ΘdH2Cp =

∨i∈J

ΘiH2Cn and ΘmH

2Cq =

⋂i∈J

ΘiH2Cn .

The greatest common right inner divisor Θ′d and the least common right inner multiple Θ′m of thefamily Θi ∈ H∞Mn

: i ∈ J are the inner functions defined by

Θ′dH2Cr =

∨i∈J

ΘiH2Cn and Θ′mH

2Cs =

⋂i∈J

ΘiH2Cn .

The Beurling-Lax-Halmos Theorem guarantees that Θd and Θm are unique up to a unitary constantright factor, and Θ′d and Θ′m are unique up to a unitary constant left factor. We write

Θd = GCD` Θi : i ∈ J, Θm = LCM` Θi : i ∈ J,Θ′d = GCDr Θi : i ∈ J, Θ′m = LCMr Θi : i ∈ J.

If n = 1, then GCD` · = GCDr · (simply denoted GCD ·) and LCM` · = LCMr · (simplydenoted LCM ·). In general, it is not true that GCD` · = GCDr · and LCM` · = LCMr ·.

However, we have:

Lemma 2.1. Let Θi := θiIn for an inner function θi (i ∈ J).


(a) GCD` Θi : i ∈ J = GCDr Θi : i ∈ J = θdIn, where θd = GCD θi : i ∈ J.(b) LCM` Θi : i ∈ J = LCMr Θi : i ∈ J = θmIn, where θm = LCM θi : i ∈ J.

Proof. (a) If Θd = GCD`Θi : i ∈ J, then

ΘdH2Cn =

∨i∈J

ΘiH2Cn =

n⊕j=1

∨i∈J

θiH2 =

n⊕j=1

θdH2,

which implies that Θd = θdIn with θd = GCD θi : i ∈ J. If instead Θd = GCDrΘi : i ∈ Jthen Θd = GCD`Θi : i ∈ J. Thus we have Θd = θdIn and hence, Θd = θdIn.

(b) If Θm = LCM`Θi : i ∈ J, then

ΘmH2Cn =

⋂i∈J

ΘiH2Cn =

n⊕j=1

⋂i∈J

θiH2 =

n⊕j=1

θmH2,

which implies that Θm = θmIn with θm = LCM θi : i ∈ J. If instead Θm = LCMrΘi : i ∈ J,then the same argument as in (a) gives the result.

In view of Lemma 2.1, if Θi = θiIn for an inner function θi (i ∈ J), we can define the greatestcommon inner divisor Θd and the least common inner multiple Θm of the Θi by

Θd ≡ GCD Θi : i ∈ J := GCD` Θi : i ∈ J = GCDr Θi : i ∈ J;Θm ≡ LCM Θi : i ∈ J := LCM` Θi : i ∈ J = LCMr Θi : i ∈ J :

they are both diagonal matrices.

For Φ ∈ L∞Mnwe write

Φ+ := Pn(Φ) ∈ H2Mn

and Φ− :=[P⊥n (Φ)

]∗ ∈ H2Mn

.

Thus we can write Φ = Φ∗− + Φ+ . Suppose Φ = [ϕij ] ∈ L∞Mnis such that Φ∗ is of bounded type.

Then we may write ϕij = θijbij , where θij is an inner function and θij and bij are coprime. Thusif θ is the least common inner multiple of θij ’s then we can write

Φ = [ϕij ] = [θijbij ] = [θaij ] = ΘA∗ (Θ ≡ θIn, A ≡ [aij ] ∈ H2Mn

). (2.6)

We note that in the factorization (2.6), A(α) is nonzero whenever θ(α) = 0. Let Φ = Φ∗− + Φ+ ∈L∞Mn

be such that Φ and Φ∗ are of bounded type. Then in view of (2.6) we can write

Φ+ = Θ1A∗ and Φ− = Θ2B

∗,

where Θi = θiIn with an inner function θi for i = 1, 2 and A,B ∈ H2Mn

. In particular, if Φ ∈ L∞Mn

is rational then the θi can be chosen as finite Blaschke products, as we observed in (2.4).By contrast with scalar-valued functions, in (2.6) Θ and A need not be (right) coprime: for

instance, if Φ := [ z zz z ] then we can write

Φ = ΘA∗ =

[z 00 z

] [1 11 1

],

but Θ := [ z 00 z ] and A := [ 1 1

1 1 ] are not right coprime because 1√2

[z −z1 1

]is a common right inner

divisor, i.e.,

Θ =1√2

[1 z−1 z

]· 1√

2

[z −z1 1

]and A =

√2

[0 10 1

]· 1√

2

[z −z1 1

]. (2.7)

If Ω = GCD` A,Θ in the representation (2.6):

Φ = ΘA∗ = A∗Θ (Θ ≡ θIn for an inner function θ),

then Θ = ΩΩ` and A = ΩA` for some inner matrix Ω` (where Ω` ∈ H2Mn

because det Θ is not

identically zero) and some Al ∈ H2Mn

. Therefore if Φ∗ ∈ L∞Mnis of bounded type then we can

writeΦ = A`

∗Ω`, where A` and Ω` are left coprime. (2.8)


A∗`Ω` is called the left coprime factorization of Φ; similarly, we can write

Φ = ΩrA∗r , where Ar and Ωr are right coprime. (2.9)

In this case, ΩrA∗r is called the right coprime factorization of Φ.

Remark 2.2. ([GHR, Corollary 2.5]) As a consequence of the Beurling-Lax-Halmos Theorem, wecan see that

Φ = ΩrA∗r (right coprime factorization) ⇐⇒ kerHΦ∗ = ΩrH

2Cn . (2.10)

In fact, if Φ = ΩrA∗r (right coprime factorization) then it is evident that

kerHΦ∗ ⊇ ΩrH2Cn .

From the Beurling-Lax-Halmos Theorem,

kerHΦ∗ = ΘH2Cn ,

for some inner function Θ, and hence (I − P )(Φ∗Θ) = 0, i.e., Φ∗ = DΘ∗, for some D ∈ H2Cn . We

want to show that Ωr = Θ up to a unitary constant right factor. Since ΘH2Cn ⊇ ΩrH

2Cn , we have

(cf. [FF, p.240]) that Ωr = Θ∆ for some square inner function ∆. Thus,

DΘ∗ = Φ∗ = ArΩ∗r = Ar∆

∗Θ∗,

which implies Ar = D∆, so that ∆ is a common right inner divisor of both Ar and Ωr. But sinceAr and Ωr are right coprime, ∆ must be a unitary constant. The proof of the converse implicationis entirely similar.

From now on, for notational convenience we write

Iω := ω In (ω ∈ H2) and H20 := IzH

2Mn

.

It is not easy to check the condition “B and Θ are coprime” in the decomposition F = B∗Θ(Θ ≡ Iθ is inner and B ∈ H2

Mn). But if F is rational (and hence Θ is given in a form Θ ≡ Iθ with

a finite Blaschke product θ) then we can obtain a more tractable criterion. To see this, we needto recall the notion of finite Blaschke-Potapov product.

Let λ ∈ D and write

bλ(z) :=z − λ1− λz

,

which is called a Blaschke factor. If M is a closed subspace of Cn then the matrix function of theform

bλPM + (I − PM ) (PM :=the orthogonal projection of Cn onto M)

is called a Blaschke-Potapov factor ; an n×n matrix function D is called a finite Blaschke-Potapovproduct if D is of the form

D = ν

M∏m=1

(bmPm + (I − Pm)

),

where ν is an n × n unitary constant matrix, bm is a Blaschke factor, and Pm is an orthogonalprojection in Cn for each m = 1, · · · ,M . In particular, a scalar-valued function D reduces to

a finite Blaschke product D = ν∏Mm=1 bm, where ν = eiω. It is also known (cf. [Po]) that an

n × n matrix function D is rational and inner if and only if it can be represented as a finiteBlaschke-Potapov product.

Write Z(θ) for the set of zeros of an inner function θ. We then have:

Lemma 2.3. Let B ∈ H∞Mnbe rational and Θ = Iθ with a finite Blaschke product θ. Then the

following statements are equivalent:

(a) B(α) is invertible for each α ∈ Z(θ);(b) B and Θ are right coprime;(c) B and Θ are left coprime.

Proof. See [CHL, Lemma 3.10].


If Θ ∈ H∞Mnis an inner matrix function, we write

H(Θ) := (ΘH2Cn)⊥;

HΘ := (ΘH2Mn

)⊥;

KΘ := (H2Mn

Θ)⊥.

If Θ = Iθ for an inner function θ then HΘ = KΘ and if n = 1, then H(Θ) = HΘ = KΘ .

The following lemma is useful in the sequel.

Lemma 2.4. If Θ ∈ H2Mn

is an inner matrix function then

dimH(Θ) <∞ ⇐⇒ Θ is a finite Blaschke-Potapov product.

Proof. Let

δ := GCD ω : ω is inner,Θ is a left inner divisor of Ω = Iω and ∆ := Iδ ,

in other words, δ is a ‘minimal’ inner function such that ∆ ≡ Iδ = ΘΘ1 for some inner matrixfunction Θ1. Note that

Θ is a finite Blaschke-Potapov product =⇒ δ is a finite Blaschke product

=⇒ dimH(∆) <∞.Observe that

H(∆) = H(ΘΘ1) = H(Θ)⊕

ΘH(Θ1) .

Thus if Θ is a finite Blaschke-Potapov product, then dimH(Θ) < ∞. Conversely, we supposedimH(Θ) <∞. Write Θ := [θij ]

nij=1. Since

rankH∗θij≤ rankH∗Θ∗ = dimH(Θ) <∞ ,

it follows that θij ’s are rational functions. Thus Θ is a rational inner matrix function and hencea finite Blaschke-Potapov product.

Lemma 2.4 implies that every inner divisor of a rational inner function (i.e., a finite Blaschke-Potapov product) is also a finite Blaschke-Potapov product: indeed, if Θ is a finite Blaschke-Potapov product and Θ1 is an inner divisor of Θ, then dimH(Θ1) ≤ dimH(Θ) < ∞, and henceby Lemma 2.4, Θ1 is a finite Blaschke-Potapov product.

From Lemma 2.4, we know that every inner divisor of Bλ := Ibλ ∈ H∞Mnis a finite Blaschke-

Potapov product. However we can say more:

Lemma 2.5. Every inner divisor of Bλ := Ibλ ∈ H∞Mnis a Blaschke-Potapov factor.

Proof. Suppose D is an inner divisor of Bλ. By Lemma 2.4, D is a Blaschke-Potapov product ofthe form

D = ν

m∏i=1

Di with Di := bλiPi + (I − Pi) (m ≤ n).

We write Bλ = ED for some E ∈ H2Mn

. Observe that Bλ(λi) is not invertible, so that λi = λ forall i = 1, 2 . . . ,m. We thus have

Pm + bλ(I − Pm) = BλD∗m = E · ν

m−1∏i=1

Di .

Then we havekerPm = ker (BλD

∗m)(λ) ⊇ kerDm−1(λ) = ranPm−1,

which implies that PmPm−1 = 0, and hence Pm and Pm−1 are orthogonal. Thus Dm−1Dm is aBlaschke-Potapov factor. Now an induction shows that D is a Blaschke-Potapov factor.


By the aid of Lemma 2.5, we can show that the equivalence (b)⇔(c) in Lemma 2.3 fails if Θis not a constant diagonal matrix. To see this, let

Θ1 :=

[bα 00 1

]and Θ2 :=

1√2

[z −z1 1

].

Then Θ := Θ1Θ2 and Θ1 are not left coprime. Observe that

Θ1 :=

[bα 00 1

], Θ =

1√2

[zbα 1−zbα 1

].

Since every right inner divisor ∆ of Θ1 is an inner divisor of Bα := Ibα , it follows from Lemma

2.5 that ∆ = Θ1 (up to a unitary constant right factor). Suppose that Θ and Θ1 are not right

coprime. Then Θ and Θ1 are not left coprime and hence Θ1 is a left inner divisor of Θ. Write

1√2

[zbα 1−zbα 1

]=

[bα 00 1

] [f11 f12

f21 f21

](fij ∈ H2).

Then we have 1√2

= bαf12, so that f12 = 1√2bα /∈ H2, giving a contradiction.

For X a subspace of H2Mn

, we write PX for the orthogonal projection from H2Mn

onto X .

Lemma 2.6. Let Θ ∈ H∞Mnbe an inner matrix function and A ∈ H2

Mn. Then the following hold:

(a) A ∈ KΘ ⇐⇒ ΘA∗ ∈ H20 ;

(b) A ∈ HΘ ⇐⇒ A∗Θ ∈ H20 ;

(c) PH20(ΘA∗) = Θ

(PKΘ

A)∗

.

Proof. Let C ∈ H2Mn

be arbitrary. We then have

A ∈ KΘ ⇐⇒ 〈A,CΘ〉 = 0

⇐⇒∫T

tr(

(CΘ)∗A)dµ = 0

⇐⇒∫T

tr (C∗AΘ∗) dµ = 0 (since tr(AB) = tr(BA))

⇐⇒ 〈AΘ∗, C〉 = 0

⇐⇒ ΘA∗ ∈ H20 ,

giving (a) and similarly, (b). For (c), we write A = A1 + A2, where A1 := PKΘA and A2 := A3Θfor some A3 ∈ H2

Mn. We then have

PH20(ΘA∗) = PH2

0(ΘA∗1 + ΘA∗2) = PH2

0

(Θ(PKΘ

A)∗ + ΘΘ∗A∗3

)= Θ(PKΘ

A)∗ ,

giving (c).

We next review the classical Hermite-Fejer interpolation problem, following [FF]; this ap-proach will be useful in the sequel. Let θ be a finite Blaschke product of degree d:

θ = eiξN∏i=1

(bi)mi

(bi :=

z − αi1− αiz

, where αi ∈ D),

where d =∑Ni=1mi. For our purposes rewrite θ in the form

θ = eiξd∏j=1

bj ,


where

bj := bk if

k−1∑l=0

ml < j ≤k∑l=0

ml

and, for notational convenience, m0 := 0. Let

ϕj :=qj

1− αjzbj−1bj−2 · · · b1 (1 ≤ j ≤ d), (2.11)

where ϕ1 := q1(1 − α1z)−1 and qj := (1 − |αj |2)

12 (1 ≤ j ≤ d). It is well known (cf. [Ta]) that

ϕjdj=1 is an orthonormal basis for H(θ).

For our purposes we concentrate on the data given by sequences of n× n complex matrices.Given the sequence Kij : 1 ≤ i ≤ N, 0 ≤ j < mi of n×n complex matrices and a set of distinctcomplex numbers α1, . . . , αN in D, the classical Hermite-Fejer interpolation problem entails findingnecessary and sufficient conditions for the existence of a contractive analytic matrix function K inH∞Mn

satisfying

K(j)(αi)

j!= Ki,j (1 ≤ i ≤ N, 0 ≤ j < mi). (2.12)

To construct a matrix polynomial K(z) ≡ P (z) satisfying (2.12), let pi(z) be the polynomial oforder d−mi defined by

pi(z) :=

N∏k=1,k 6=i

( z − αkαi − αk

)mk.

Consider the matrix polynomial P (z) of degree d− 1 defined by

P (z) :=

N∑i=1

(K ′i,0 +K ′i,1(z − αi) +K ′i,2(z − αi)2 + · · ·+K ′i,mi−1(z − αi)mi−1

)pi(z), (2.13)

where the K ′i,j are obtained by the following equations:

K ′i,j = Ki,j −j−1∑k=0

K ′i,k p(j−k)i (αi)

(j − k)!(1 ≤ i ≤ N ; 0 ≤ j < mi)

and K ′i,0 = Ki,0 (1 ≤ i ≤ N). Then P (z) satisfies (2.12).

On the other hand, for an inner function θ, let Uθ be defined by the compression of the shiftoperator U : i.e.,

Uθ = PH(θ)U |H(θ).

Let Θ = Iθ and W be the unitary operator from⊕d

1 Cn onto H(Θ) defined by

W := (Iϕ1, Iϕ2

, · · · , Iϕd), (2.14)

where the ϕj are the functions in (2.11). It is known [FF, Theorem X.1.5] that if θ is the finiteBlaschke product of order d, then Uθ is unitarily equivalent to the lower triangular matrix M onCd defined by

M :=

α1 0 0 0 · · · · · ·q1q2 α2 0 0 · · · · · ·

−q1α1q3 q2q3 α3 0 · · · · · ·q1α2α3q4 −q2α3q4 q3q4 α4 · · · · · ·

−q1α2α3α4q5 q2α3α4q5 −q3α4q5 q4q5 α5

. . .

......

.... . .

. . .. . .

. (2.15)


If L ∈Mn and M = [mi,j ]d×d, then the matrix L⊗M is the matrix on Cn×d defined by the blockmatrix

L⊗M :=

Lm1,1 Lm1,2 · · · Lm1,d

Lm2,1 Lm2,2 · · · Lm2,d

......

......

Lmd,1 Lmd,2 · · · Lmd,d

.Now let P (z) ∈ H∞Mn

be a matrix polynomial of degree k. Then the matrix P (M) on Cn×d isdefined by

P (M) :=

k∑i=0

Pi ⊗M i, where P (z) =

k∑i=0

Pizi. (2.16)

For Φ ∈ H∞Mnand Θ := Iθ with an inner function θ, we write, for brevity,

(TΦ)Θ := PH(Θ)TΦ|H(Θ), (2.17)

which is called the compression of TΦ to H(Θ). If M is given by (2.15) and P is the matrixpolynomial defined by (2.13) then the matrix P (M) is called the Hermite-Fejer matrix determinedby (2.16). In particular, it is known [FF, Theorem X.5.6] that

W ∗(TP )ΘW = P (M), (2.18)

which says that P (M) is a matrix representation for (TP )Θ.

Lemma 2.7. Let A ∈ H∞Mnand Θ = Iθ for a finite Blaschke product θ. If A(α) is invertible for all

α ∈ Z(θ) then (TA)Θ is invertible.

Proof. Suppose (TA)Θf = 0 for some f ∈ H(Θ), so that PH(Θ)(Af) = 0 and hence, Af ∈ ΘH2Cn .

Since A(α) is invertible for all α ∈ Z(θ), it follows that f ∈ ΘH2Cn and hence, f ∈ ΘH2

Cn ∩H(Θ) =0. Thus (TA)Θ is one-one. But since (TA)Θ is a finite dimensional operator (because θ is a finiteBlaschke product), it follows that (TA)Θ is invertible.

3. Hyponormality of Block Toeplitz Operators

To get a tractable criterion for the hyponormality of block Toeplitz operators with bounded typesymbols, we need a triangular representation for compressions of the unilateral shift operatorU ≡ Tz. We refer to [AC] and [Ni] for details on this representation. For an explicit criterion,we need to introduce the triangularization theorem concretely. To do so, recall that for an innerfunction θ, Uθ is defined by

Uθ = PH(θ)U |H(θ). (3.1)

There are three cases to consider.

Case 1 : Let B be a Blaschke product and let Λ := λn : n ≥ 1 be the sequence of zeros of Bcounted with their multiplicities. Write

β1 := 1, βk :=

k−1∏n=1

λn − z1− λnz

· |λn|λn

(k ≥ 2),

and let

δj :=dj

1− λjzβj (j ≥ 1),


where dj := (1− |λj |2)12 . Let µB be a measure on N given by µB(n) := 1

2d2n, (n ∈ N). Then the

map VB : L2(µB)→ H(B) defined by

VB(c) :=1√2

∑n≥1

c(n)dnδn, c ≡ c(n)n≥1, (3.2)

is unitary and UB is mapped onto the operator

V ∗BUBVB = (I − JB)MB , (3.3)

where (MBc)(n) := λnc(n) (n ∈ N) is a multiplication operator and

(JBc)(n) :=

n−1∑k=1

c(k)|λk|−2 · βn(0)

βk(0)dkdn (n ∈ N)

is a lower-triangular Hilbert-Schmidt operator.

Case 2 : Let s be a singular inner function with continuous representing measure µ ≡ µs. Let µλbe the projection of µ onto the arc ζ : ζ ∈ T, 0 < argζ ≤ argλ and let

sλ(ζ) := exp(−∫T

t+ ζ

t− ζdµλ(t)

)(ζ ∈ D).

Then the map Vs : L2(µ)→ H(s) defined by

(Vsc)(ζ) =√

2

∫Tc(λ)sλ(ζ)

λdµ(λ)

λ− ζ(ζ ∈ D) (3.4)

is unitary and Us is mapped onto the operator

V ∗s UsVs = (I − Js)Ms, (3.5)

where (Msc)(λ) := λc(λ) (λ ∈ T) is a multiplication operator and

(Jsc)(λ) = 2

∫Teµ(t)−µ(λ)c(t)dµλ(t) (λ ∈ T)


Case 3 : Let ∆ be a singular inner function with pure point representing measure µ ≡ µ∆. Weenumerate the set t ∈ T : µ(t) > 0 as a sequence tkk∈N. Write µk := µ(tk), k ≥ 1.Further, let µ∆ be a measure on R+ = [0,∞) such that dµ∆(λ) = µ[λ]+1dλ and define a function∆λ on the unit disk D by the formula

∆λ(ζ) := exp

−

[λ]∑k=1

µktk + ζ

tk − ζ− (λ− [λ])µ[λ]+1

t[λ]+1 + ζ

t[λ]+1 − ζ

,

where [λ] is the integer part of λ (λ ∈ R+) and by definition ∆0 := 1. Then the map V∆ :L2(µ∆)→ H(∆) defined by

(V∆c)(ζ) :=√

2

∫R+

c(λ)∆λ(ζ)(1− t[λ]+1ζ)−1dµ∆(λ) (ζ ∈ D) (3.6)

is unitary and U∆ is mapped onto the operator

V ∗∆U∆V∆ = (I − J∆)M∆, (3.7)

where (M∆c)(λ) := t[λ]+1c(λ), (λ ∈ R+) is a multiplication operator and

(J∆c)(λ) := 2

∫ λ

0

c(t)∆λ(0)

∆t(0)dµ∆(t) (λ ∈ R+)


Collecting the above three cases we get:


Triangularization theorem. ([Ni, p.123]) Let θ be an inner function with the canonical factorizationθ = B · s ·∆, where B is a Blaschke product, and s and ∆ are singular functions with representingmeasures µs and µ∆ respectively, with µs continuous and µ∆ a pure point measure. Then the mapV : L2(µB)× L2(µs)× L2(µ∆)→ H(θ) defined by

V :=

VB 0 00 BVs 00 0 BsV∆

(3.8)

is unitary, where VB , µB , VS , µS , V∆, µ∆ are defined in (3.2) - (3.7) and Uθ is mapped onto theoperator

M := V ∗UθV =

MB 0 00 Ms 00 0 M∆

+ J,

where MB ,MS ,M∆ are defined in (3.3), (3.5) and (3.7) and

J := −

JBMB 0 00 JsMs 00 0 J∆M∆

+A

is a lower-triangular Hilbert-Schmidt operator, with A3 = 0, rankA ≤ 3.

If Φ ∈ L∞Mn, then by (1.3),

[T ∗Φ, TΦ] = H∗Φ∗HΦ∗ −H∗ΦHΦ + TΦ∗Φ−ΦΦ∗ .

Since the normality of Φ is a necessary condition for the hyponormality of TΦ, the positivity ofH∗Φ∗HΦ∗ − H∗ΦHΦ is an essential condition for the hyponormality of TΦ. Thus, we isolate thisproperty as a new notion, weaker than hyponormality. The reader will notice at once that thisnotion is meaningful for non-scalar symbols.

Definition 3.1. Let Φ ∈ L∞Mn. The pseudo-selfcommutator of TΦ is defined by

[T ∗Φ, TΦ]p := H∗Φ∗HΦ∗ −H∗ΦHΦ.

TΦ is said to be pseudo-hyponormal if [T ∗Φ, TΦ]p is positive semidefinite.

As in the case of hyponormality of scalar Toeplitz operators, we can see that the pseudo-hyponormality of TΦ is independent of the constant matrix term Φ(0). Thus whenever we considerthe pseudo-hyponormality of TΦ we may assume that Φ(0) = 0. Observe that if Φ ∈ L∞Mn

then

[T ∗Φ, TΦ] = [T ∗Φ, TΦ]p + TΦ∗Φ−ΦΦ∗ .

We thus have

TΦ is hyponormal ⇐⇒ TΦ is pseudo-hyponormal and Φ is normal;

and (via [GHR, Theorem 3.3]) TΦ is pseudo-hyponormal if and only if E(Φ) 6= ∅.

For Φ ≡ Φ∗− + Φ+ ∈ L∞Mn, we write

C(Φ) :=K ∈ H∞Mn

: Φ−KΦ∗ ∈ H∞Mn

.

Thus if Φ ∈ L∞Mnthen

K ∈ E(Φ) ⇐⇒ K ∈ C(Φ) and ||K||∞ ≤ 1.

Also if K ∈ C(Φ) then HΦ∗−= HKΦ∗+

= T ∗KHΦ∗+

, which gives a necessary condition for the

nonempty-ness of C(Φ) (and hence the hyponormality of TΦ): in other words,

K ∈ C(Φ) =⇒ kerHΦ∗+⊆ kerHΦ∗−

. (3.9)

We begin with:


Proposition 3.2. Let Φ ≡ Φ∗− + Φ+ ∈ L∞Mnbe such that Φ and Φ∗ are of bounded type. Thus we

may write

Φ+ = Θ1A∗ and Φ− = Θ2B

∗,

where Θi = Iθi for an inner function θi (i = 1, 2) and A,B ∈ H2Mn

. If C(Φ) 6= ∅, then Θ2 is aninner divisor of Θ1, i.e., Θ1 = Θ0Θ2 for some inner function Θ0.

Proof. In view of (2.6) we may write

Φ+ ≡ [θ1aij ]n×n and Φ− =[θ2bij

]n×n = [θijcij ]n×n ,

where each θij is an inner function, cij ∈ H2, θij and cij are coprime, and θ2 is the least commonmultiple of θij ’s. Suppose C(Φ) 6= ∅. Then there exists a matrix function K ∈ H∞Mn

such that

Φ∗− −KΦ∗+ ∈ H2Mn

. Thus BΘ∗2 −KAΘ∗1 ∈ H2Mn

, which implies that

BΘ∗2Θ1 =[bjiθ2θ1

]∈ H2

Mn.

But since θ2bij = θijcij , and hence bij = θ2θijcij , it follows that

bjiθ2θ1 =(θ2θjicji

)θ2θ1 = θjicjiθ1 ∈ H∞.

Since θji and cji are coprime, we have that

θjiθ1 ∈ H∞, and hence θ2θ1 ∈ H∞,

which implies that Θ2 divides Θ1.

Proposition 3.2 shows that the hyponormality of Tϕ with scalar-valued rational symbol ϕimplies

deg (ϕ−) ≤ deg (ϕ+),

which is a generalization of the well-known result for the cases of the trigonometric Toeplitz

operators, i.e., if ϕ =∑Nn=−m anz

n is such that Tϕ is hyponormal then m ≤ N (cf. [FL]).

In view of Proposition 3.2, when we study the hyponormality of block Toeplitz operatorswith bounded type symbols Φ (i.e., Φ and Φ∗ are of bounded type) we may assume that the symbolΦ ≡ Φ∗− + Φ+ ∈ L∞Mn

is of the form

Φ+ = Θ1Θ0A∗ and Φ− = Θ1B

∗,

where Θi := Iθi for an inner function θi (i = 0, 1) and A,B ∈ H2Mn

.

Our criterion is as follows:

Theorem 3.3. Let Φ ≡ Φ∗− + Φ+ ∈ L∞Mnbe normal such that Φ and Φ∗ are of bounded type of the

form


∗,

where Θi = Iθi for an inner function θi (i = 0, 1) and A,B ∈ H2Mn

. WriteV : L ≡ L2(µB)× L2(µs)× L2(µ∆)→ H(θ1θ0) is unitary as in (3.8);

M := V ∗Uθ1θ0V ;

L := L⊗ Cn

V := V ⊗ In.

(3.10)

If K ∈ C(Φ) then

[T ∗Φ, TΦ] = (TA)∗Θ1Θ0V(I|L −K(M)∗K(M)

)V∗(TA)Θ1Θ0

⊕0|Θ1Θ0H2

Cn,

where K(M) is understood as an H∞-functional calculus. Hence, in particular,

K(M) is contractive =⇒ TΦ is hyponormal;


the converse is also true if (TA)Θ1Θ0 has dense range, and in this case,

rank [T ∗Φ, TΦ] = rank(I|L −K(M)∗K(M)

).

Proof. Let E,F ∈ H(Θ1Θ0) and K ∈ C(Φ). Since kerHΘ∗1Θ∗0= Θ1Θ0H

2Cn , we have

HΘ∗1Θ∗0KE = HΘ∗1Θ∗0

(PH(Θ1Θ0)(KE)).

Since H∗Θ∗1Θ∗0HΘ∗1Θ∗0

is the projection onto H(Θ1Θ0), it follows that⟨H∗Θ∗1Θ∗0K

HΘ∗1Θ∗0KE, F

⟩=⟨PH(Θ1Θ0)KE, PH(Θ1Θ0)KF

⟩=⟨

(TK)Θ1Θ0E, (TK)Θ1Θ0

F⟩,

which gives

H∗Θ∗1Θ∗0KHΘ∗1Θ∗0K

|H(Θ1Θ0) = (TK)∗Θ1Θ0(TK)Θ1Θ0

.

Observe that [T ∗Φ, TΦ] = H∗AΘ∗0Θ∗1HAΘ∗0Θ∗1

−H∗BΘ∗1HBΘ∗1

because Φ is normal. But since

cl ran (H∗AΘ∗0Θ∗1HAΘ∗0Θ∗1

) =(kerHAΘ∗0Θ∗1

)⊥ ⊆ (Θ1Θ0H2Cn)⊥ = H(Θ1Θ0)

and

cl ran (H∗BΘ∗1HBΘ∗1

) =(kerHBΘ∗1

)⊥ ⊆ (Θ1H2Cn)⊥ = H(Θ1),

we have ran [T ∗Φ, TΦ] ⊆ H(Θ1Θ0). But since Φ−KΦ∗ ∈ H∞Mn, it follows that on H(Θ1Θ0),

[T ∗Φ, TΦ] =(H∗Φ∗HΦ∗ −H∗ΦHΦ

)∣∣∣H(Θ1Θ0)

=(H∗Φ∗HΦ∗ −H∗KΦ∗HKΦ∗

)∣∣∣H(Θ1Θ0)

=(H∗Θ∗1Θ∗0A+Φ−HΘ∗1Θ∗0A+Φ− −H∗K(Θ∗1Θ∗0A+Φ−)HK(Θ∗1Θ∗0A+Φ−)

)∣∣∣H(Θ1Θ0)

= T ∗(A+Θ1Θ0Φ−)

(H∗Θ∗1Θ∗0

HΘ∗1Θ∗0−H∗KΘ∗1Θ∗0

HKΘ∗1Θ∗0

)T(A+Θ1Θ0Φ−)

∣∣∣H(Θ1Θ0)

= (TA)∗Θ1Θ0

(I|H(Θ1Θ0) − (TK)∗Θ1Θ0

(TK)Θ1Θ0

)(TA)Θ1Θ0

,

where (TA)Θ1Θ0 is understood in the sense that the compression (TA)Θ1Θ0 is bounded even thoughTA is possibly unbounded; in fact,

PH(Θ1Θ0)T(A+Θ1Θ0Φ−)|H(Θ1Θ0) = PH(Θ1Θ0)TA|H(Θ1Θ0).

On the other hand, since K(z) ≡[krs(z)

]1≤r,s≤n ∈ H

∞Mn

, we may write

K(z) =

∞∑i=0

Kizi (Ki ∈Mn).

We also write krs(z) :=∑∞

0 c(rs)i zi and then Ki =

[c(rs)i

]1≤r,s≤n

. We thus have


(TK)Θ1Θ0 = PH(Θ1Θ0)TK |H(Θ1Θ0)

=[PH(θ1θ0)Tkrs |H(θ1θ0)

]1≤r,s≤n

=

[ ∞∑i=0

c(rs)i PH(θ1θ0)Tzi |H(θ1θ0)

]1≤r,s≤n

=

[ ∞∑i=0

c(rs)i

(PH(θ1θ0)Tz|H(θ1θ0)

)i]1≤r,s≤n

(because θ1θ0H2 ⊆ LatTz)

=

∞∑i=0

(U iθ1θ0 ⊗

[c(rs)i

]1≤r,s≤n

)

=

∞∑i=0

(U iθ1θ0 ⊗ Ki

),

Let φj be an orthonormal basis forH(θ1θ0) and put ej := V ∗φj . Then ej forms an orthonormalbasis for L2(µB)× L2(µs)× L2(µ∆). Thus for each f ∈ Cn, we have V(ej ⊗ f) = φj ⊗ f . It thusfollows that ⟨

(TK)Θ1Θ0(φj ⊗ f) , φk ⊗ g

⟩=

∞∑i=0

⟨(U iθ1θ0 ⊗Ki)(φj ⊗ f), φk ⊗ g

⟩=

∞∑i=0

⟨(U iθ1θ0φj)⊗ (Kif), φk ⊗ g

⟩=

∞∑i=0

⟨U iθ1θ0V ej , V ek

⟩⟨Kif, g

⟩=

∞∑i=0

⟨(M i ⊗Ki)(ej ⊗ f), ek ⊗ g

⟩=⟨K(M)(ej ⊗ f) , ek ⊗ g

⟩,

which implies that

V∗(TK)Θ1Θ0V = K(M). (3.11)

Here K(M) is understood as a H∞-functional calculus (so called the Sz.-Nagy-Foias functionalcalculus) because M is an absolutely continuous contraction: in fact, we claim that

every compression of the shift operator is completely non-unitary.

To see this, write PXUPX for the compression of U to X ≡ PXH2 with some projection PX .

We assume to the contrary that PXUPX has a unitary summand W acting on a closed subspaceY ⊆ X . But since ||U || = 1, we must have that PY⊥U |Y = 0. Thus we can see that Y is aninvariant subspace of U . Thus, by Beurling’s Theorem, Y = θH2 for some inner function θ. Butthen W (θH2) = zθH2, and hence W is not surjective, a contradiction. Hence every compressionof the shift operator is completely non-unitary. We can therefore conclude that

[T ∗Φ, TΦ] = (TA)∗Θ1Θ0V(I|L −K(M)∗K(M)

)V∗(TA)Θ1Θ0

⊕0|Θ1Θ0H2

Cn.

The remaining assertions follow trivially from the first assertion.

If Φ is a scalar-valued function then Theorem 3.3 reduces to the following corollary.

Corollary 3.4. Let ϕ ≡ ϕ∗− + ϕ+ ∈ L∞ be such that ϕ and ϕ are of bounded type of the form

ϕ+ = θ1θ0a and ϕ− = θ1b,


where θ1 and θ0 are inner functions and a, b ∈ H2. If k ∈ C(ϕ) then

Tϕ is hyponormal ⇐⇒ k(M) is contractive,

where M is defined as in (3.10).

Proof. By Theorem 3.3, it suffices to show that (Ta)θ1θ0 has dense range. To prove this suppose(Ta)∗θ1θ0f = 0 for some f ∈ H(θ1θ0). Then PH(θ1θ0)(af) = 0, i.e., af = θ1θ0h for some h ∈ H2.

Thus we have aθ1θ0f ∈(H2)⊥ ∩H2 = 0, which implies that f = 0. Therefore (Ta)∗θ1θ0 is 1-1,

which gives the result.

Remark 3.5. We note that in Corollary 3.4, if ϕ is a rational function then M is a finite matrix.Indeed, if ϕ is a rational function, and hence θ1θ0 is a finite Blaschke product of the form

θ1θ0 =

d∏j=1

z − αj1− αjz

,

then M is obtained by (2.15).

Remark 3.6. We mention that K ∈ C(Φ) may not be contractive, i.e., there might exist a functionK ∈ C(Φ) \ E(Φ). In spite of this, Theorem 3.3 guarantees that

I −K(M)∗K(M)

is unchanged regardless of the particular choice of K in C(Φ). We will illustrate this phenomenonwith a scalar-valued Toeplitz operator with a trigonometric polynomial symbol. For example, let

Φ(z) = z−2 + 2z−1 + z + 2z2.

If K(z) = 12 + 3

4z, then Φ∗− −KΦ∗+ = (z−2 + 2z−1)− ( 12 + 3

4z)(z−1 + 2z−2) = − 3

4 ∈ H∞, so that

K ∈ C(Φ), but ||K||∞ = 54 > 1. However by (2.15) we have

M =

[0 01 0

]and hence, K(M) =

[12 034

12

],

so that

I −K(M)∗K(M) =

[1 00 1

]−[

1316

38

38

14

]=

[316 − 3

8− 3

834

]≥ 0,

which implies that TΦ is hyponormal even though ||K||∞ > 1. Of course, in view of Cowen’s

Theorem, there exists a function b ∈ E(Φ): indeed, b(z) =z+ 1

2

1+ 12 z∈ E(Φ).

We now provide some revealing examples that illustrate Theorem 3.3.

Example 3.7. Let ∆ be a singular inner function of the form

∆ := exp(z + 1

z − 1

)and consider the matrix-valued function

Φ :=

[∆ z∆ + z∆

z∆ + z∆ ∆

].

We now use Theorem 3.3 to determine the hyponormality of TΦ. Under the notation of Theorem3.3 we have

Θ1 = Iz∆, Θ0 = I2, A =

[0 11 0

], B =

[z 11 z

].

If we put

K(z) :=

[1 zz 1

],


then a straightforward calculation shows that K ∈ C(Φ). Under the notation of Theorem 3.3 wecan see that

(TA)Iz∆ =

[0 I|LI|L 0

]: L → L is invertible.

But since

K(M) =

[I|L MM I|L

],

it follows that

I|L −K(M)∗K(M) = −[M∗M M +M∗

M +M∗ M∗M

],

which is not positive (simply by looking at the upper-left entry). It therefore follows from Theorem3.3 that TΦ is not hyponormal.

Example 3.8. Let ∆ be a singular inner function of the form

∆ := exp

(z + 1

z − 1

)and let

Ω := ∆12 = exp

(1

2· z + 1

z − 1

).

Consider the function

ϕ :=4

5z +

8

5∆ +

37

50

(z2Ω− 1√

ez2 +

1√ez)

+29

25(z + 2∆).

Observe that

ϕ− =4

5z +

8

5∆ +

37

50(I − P )(z2Ω) + c (c ∈ C).

We thus have

k :=25

29

(4

5+

37

100z2Ω

)∈ C(ϕ),

but

||k||∞ =117

116> 1.

Thus by the aid of such a function k, we cannot determine the hyponormality of Tϕ. Using thenotation in the triangularization theorem we can show that

(i) L2(µ∆) = L2(0, 1) and L2(µB) = C;(ii) M∆ = I and MB = 0;

(iii) (J∆c)(λ) = 2∫ λ

0et−λc(t)dt for λ ∈ (0, 1) and c ∈ L2(0, 1);

(iv) Vz = 1√2

and (V∆c)(ζ) =√

2∫ 1

0c(λ)exp

(−λ 1+ζ

1−ζ)(1− ζ)−1dλ, ζ ∈ D, c ∈ L2(0, 1);

(v) M ≡MB ×M∆ + J : L→ L, where L ≡ C⊕ L2(0, 1) ∼= H(Iz∆).

Note that H(z∆) = H(z)⊕ zH(∆), so that PH(z∆) = PH(z) + zPH(∆)z. We then have

Uz∆ =

[Uz 0a zU∆z

]:

[H(z)zH(∆)

]→[H(z)zH(∆)

].

Since Uz = 0, it follows that

M =

[Vz 00 zV∆

]∗ [0 0a zU∆z

] [Vz 00 zV∆

]=

[0 0

(zV∆)∗aVz I − J∆

].

By using the fact that PH(θ) = I − θPθ, we can compute:

a = PzH(∆)(z · 1) = zPH(∆)z(z · 1) = zPH(∆)(1) = z(I −∆P∆)(1) = z

(1− ∆

e

).


Thus we have

M =

[0 0

V ∗∆(1− ∆

e

)Vz I − J∆

].

Write

A := V ∗∆

(1− ∆

e

)Vz.

Then we have

||A||2 =∣∣∣∣∣∣1− ∆

e

∣∣∣∣∣∣2 =∣∣∣∣∣∣1− ∆(0)

e− 1

e

(∆−∆(0)

)∣∣∣∣∣∣2=(

1− ∆(0)

e

)2

+1

e2

(||∆||2 − |∆(0)|2

)=(

1− 1

e2

)2

+1

e2

(1− 1

e2

)= 1− 1

e2.

A straightforward calculation shows that

k(M) =25

29

[45 0

S(

45 + 37

100z2Ω)

(I − J∆)

],

where

S :=

(( 37

100zΩ)

(I − J∆)

)A.

On the other hand, consider the function

ϕ1(z) := (I − P )(zΩ) + ∆.

Then q := zΩ ∈ E(ϕ1). Since M∆ = 1, it follows from Corollary 3.4 that q(M) = (zΩ)(I − J∆) isa contraction. Thus we have

||S|| = 37

100||(zΩ)(I − J∆)A|| ≤ 37

100

√1− 1

e2

Also we consider the function

ϕ2(z) :=4

5∆ +

9

25(I − P )(z2 Ω) + ∆.

Put

B(z) :=z2Ω + 4

5

1 + 45z

2Ω.

Since B(z) = 45 + 9

25z2Ω + ∆g for some g ∈ H2, we have

(ϕ2)− − B (ϕ2)+ =4

5∆ +

9

25(I − P )(z2 Ω)−

(4

5+

9

25z2Ω + ∆g

)∆

= − 9

25P (z2Ω)− g ∈ H2,

Since ||B||∞ = 1, it follows that Tϕ2is hyponormal. In particular, since

r :=4

5+

9

25z2Ω ∈ C(ϕ2),

it follows from Corollary 3.4 that r(M) = ( 45 + 9

25z2Ω)(I − J∆) is a contraction. Thus we have

that ∣∣∣∣∣∣∣∣(4

5+

37

100z2Ω

)(I − J∆)

∣∣∣∣∣∣∣∣ ≤ ∣∣∣∣∣∣∣∣(4

5+

9

25z2Ω

)(I − J∆)

∣∣∣∣∣∣∣∣+1

100||(z2Ω)(I − J∆)|| ≤ 101

100.

Using the observation that if A,B,C and D are operators then∣∣∣∣∣∣∣∣[A BC D

]∣∣∣∣∣∣∣∣ ≤ ∣∣∣∣∣∣∣∣[||A|| ||B||||C|| ||D||

]∣∣∣∣∣∣∣∣ ,


we can see that

||k(M)|| ≤ 25

29

∣∣∣∣∣∣∣∣∣∣[

45 0

||S||∣∣∣∣∣∣( 4

5 + 925z

2Ω)

(I − J∆)∣∣∣∣∣∣] ∣∣∣∣∣∣∣∣∣∣

≤ 25

29

∣∣∣∣∣∣∣∣∣∣[

45 0

37100

√1− 1

e2101100

] ∣∣∣∣∣∣∣∣∣∣≈ 0.968 < 1,

which, by Corollary 3.4, implies that Tϕ is hyponormal.

We now consider the condition “C(Φ) 6= ∅”, i.e., the existence of a function K ∈ H∞Mnsuch

that Φ−KΦ∗ ∈ H∞Mn. In view of (3.9), we may assume that

kerHΦ∗+⊆ kerHΦ∗−

(3.12)

whenever we study the hyponormality of TΦ. Recall ([Gu1, Corollary 2]) that

H∗Φ∗HΦ∗ −H∗ΦHΦ ≥ 0 ⇐⇒ ∃K ∈ H∞Mnwith ||K||∞ ≤ 1 such that HΦ∗−

= T ∗KHΦ∗+

. (3.13)

We thus have

C(Φ) 6= ∅ ⇐⇒ ∃K ∈ H∞Mnsuch that Φ∗− −KΦ∗+ ∈ H2

Mn

⇐⇒ HΦ∗−= T ∗

KHΦ∗+

for some K ∈ H∞Mn

⇐⇒ H∗αΦ∗+HαΦ∗+

−H∗Φ∗−HΦ∗−≥ 0 for some α > 0 (by (3.13))

⇐⇒ kerHΦ∗+⊆ kerHΦ∗−

and

sup

||HΦ∗−

F ||||HΦ∗+

F ||: F ∈ ker (HΦ∗+

)⊥, ||F || = 1

≤ α.

(3.14)

If Φ ∈ L∞Mnis a rational function then by Kronecker’s lemma (cf. [Ni, p.183]), ranHΦ∗+

is finite

dimensional. Thus by (3.14) we can see that

C(Φ) 6= ∅ ⇐⇒ kerHΦ∗+⊆ kerHΦ∗−

.

Consequently, if Φ ∈ L∞Mnis a rational function then there always exists a function K ∈ C(Φ)

under the kernel assumption (3.12). We record this in

Proposition 3.9. If Φ ∈ L∞Mnis a rational function satisfying kerHΦ∗+

⊆ kerHΦ∗−, then C(Φ) 6= ∅.

We remark that there is an explicit way to find a function (in fact, a matrix-valued polynomial)K in C(Φ) for the rational symbol case. To see this, in view of Proposition 3.2, suppose Φ ∈ L∞Mn

is of the form


∗,

where Θi = Iθi for a finite Blaschke product θi (i = 0, 1). We observe first that

K ∈ C(Φ) ⇐⇒ Φ−KΦ∗ ∈ H∞Mn⇐⇒ Θ0B −KA ∈ Θ1Θ0H

∞Mn

. (3.15)

Suppose θ1θ0 is a finite Blaschke product of degree d of the form

θ1θ0 =

N∏i=1

(z − αi1− αiz

)mi(d :=

N∑i=1

mi).

Then the last assertion in (3.15) holds if and only if the following equations hold: for each i =1, . . . , N ,

Bi,0Bi,1Bi,2

...Bi,mi−1

=

Ki,0 0 0 · · · 0Ki,1 Ki,0 0 · · · 0Ki,2 Ki,1 Ki,0 · · · 0

.... . .

. . .. . .

...Ki,mi−1 Ki,mi−2 . . . Ki,1 Ki,0

Ai,0Ai,1Ai,2

...Ai,mi−1

, (3.16)


where

Ki,j :=K(j)(αi)

j!, Ai,j :=

A(j)(αi)

j!and Bi,j :=

(θ0B)(j)(αi)

j!.

Thus K is a function in H∞Mnfor which

K(j)(αi)

j!= Ki,j (1 ≤ i ≤ N, 0 ≤ j < mi), (3.17)

where the Ki,j are determined by the equation (3.16). This is exactly the classical Hermite-Fejerinterpolation problem which we have introduced in Section 2. Thus the solution (2.13) for theclassical Hermite-Fejer interpolation problem provides a polynomial K ∈ C(Φ).

Therefore we get:

Proposition 3.10. If Φ ∈ L∞Mnis a rational function such that C(Φ) 6= ∅, then C(Φ) contains a

polynomial.

However, by comparison with the rational symbol case, there may not exist a function K ∈C(Φ) if Φ is of bounded type. But we guarantee the existence of a function K ∈ C(Φ) if thebounded type symbol Φ satisfies a certain determinant property. To see this, we recall the notionof the reduced minimum modulus. If T ∈ B(H) then the reduced minimum modulus of T is definedby

γ(T ) =

inf||Tx|| : x ∈

(kerT

)⊥, ||x|| = 1

if T 6= 0

0 if T = 0.

It is well known ([Ap]) that if T 6= 0 then γ(T ) > 0 if and only if T has closed range. We caneasily show that if S, T ∈ B(H) and S is one-one then

γ(ST ) ≥ γ(S)γ(T ). (3.18)

We then have:

Proposition 3.11. Let Φ ∈ L∞Mnbe such that Φ and Φ∗ are of bounded type satisfying


.

If there exists δ > 0 such that M :=t : |det Φ+(eit)| < δ

has measure zero then C(Φ) 6= ∅.

Proof. Suppose M has measure zero for some δ > 0. Write

Φ+ = ΘA∗ (right coprime factorization).

Since det Θ is inner, we have |det Φ+| = |detA| a.e. on T. Then by the well-known result [GGK,Theorem XXIII.2.4], our determinant condition shows that the multiplication operator MA isinvertible and γ(MA) > 0, where MAf := Af for f ∈ L2

Cn . Since A ∈ H∞Mn, the Toeplitz operator

TA is a restriction of MA. Thus it follows that γ(TA) ≥ γ(MA) > 0. Since ranHΘ∗ = H(Θ), itfollows that

HΦ∗+= HAΘ∗ = T ∗

AHΘ∗ = T ∗

A|H(Θ)HΘ∗ .

Observe that

γ(HΘ∗) = inf||HΘ∗F || : F ∈ H(Θ), ||F || = 1

= inf

||Θ∗F || : F ∈ H(Θ), ||F || = 1

= 1.

We now claim that

T ∗A|H(Θ) is one-one. (3.19)

Indeed, since

ΘH2Cn = kerHAΘ∗ = kerT ∗

AHΘ∗ and kerHΘ∗ = ΘH2

Cn ,


it follows that T ∗A|ranHΘ∗ = T ∗

A|H(Θ) is one-one, which gives (3.19). Now since γ(TA) > 0 it follows

from (3.18) and (3.19) that

γ(HΦ∗) = γ(HΦ∗+) = γ(T ∗

A|H(Θ)HΘ∗) ≥ γ(T ∗

A|H(Θ)) ≥ γ(T ∗

A) = γ(TA) = γ(TA) > 0.

We thus have

sup

||HΦ∗−

F ||||HΦ∗+

F ||: F ∈

(kerHΦ∗+

)⊥, ||F || = 1

≤ ||HΦ||γ(HΦ∗)

< α for some α > 0. (3.20)

Therefore, by (3.14) we can conclude that C(Φ) 6= ∅.

4. Subnormality of Block Toeplitz Operators

As we saw in Introduction, the Bram-Halmos criterion on subnormality ([Br], [Con]) says thatT ∈ B(H) is subnormal if and only if the positive test (1.6) holds. It is easy to see that (1.6) isequivalent to the following positivity test:

I T ∗ . . . T ∗k

T T ∗T . . . T ∗kT...

.... . .

...T k T ∗T k . . . T ∗kT k

≥ 0 (all k ≥ 1). (4.1)

Condition (4.1) provides a measure of the gap between hyponormality and subnormality. In factthe positivity condition (4.1) for k = 1 is equivalent to the hyponormality of T , while subnormalityrequires the validity of (4.1) for all k. For k ≥ 1, an operator T is said to be k-hyponormal ifT satisfies the positivity condition (4.1) for a fixed k. Thus the Bram-Halmos criterion can bestated as: T is subnormal if and only if T is k-hyponormal for all k ≥ 1. The k-hyponormality hasbeen considered by many authors with an aim at understanding the gap between hyponormalityand subnormality. For instance, the Bram-Halmos criterion on subnormality indicates that 2-hyponormality is generally far from subnormality. There are special classes of operators, however,for which these two notions are equivalent. A trivial example is given by the class of operatorswhose square is compact (e.g., compact perturbations of nilpotent operators of nilpotency 2).Also in [CuL1, Example 3.1], it was shown that there is no gap between 2-hyponormality andsubnormality for back-step extensions of recursively generated subnormal weighted shifts.

On the other hand, in 1970, P.R. Halmos posed the following problem, listed as Problem 5 inhis lectures “Ten problems in Hilbert space” [Hal1], [Hal2]:

Is every subnormal Toeplitz operator either normal or analytic ?

A Toeplitz operator Tϕ is called analytic if ϕ ∈ H∞. Any analytic Toeplitz operator is easilyseen to be subnormal: indeed, Tϕh = P (ϕh) = ϕh = Mϕh for h ∈ H2, where Mϕ is the normaloperator of multiplication by ϕ on L2. The question is natural because the two classes, the normaland analytic Toeplitz operators, are fairly well understood and are subnormal. Halmos’s Problem5 has been partially answered in the affirmative by many authors (cf. [Ab], [AIW], [Co2], [Co3],[CuL1], [CuL2], [NT], and etc). In 1984, Halmos’s Problem 5 was answered in the negative byC. Cowen and J. Long [CoL]: they found an analytic function ψ for which Tψ+αψ (0 < α < 1) issubnormal - in fact, this Toeplitz operator is unitarily equivalent to a subnormal weighted shift Wβ

with weight sequence β ≡ βn, where βn = (1−α2n+2)12 for n = 0, 1, 2, . . .. Unfortunately, Cowen

and Long’s construction does not provide an intrinsic connection between subnormality and thetheory of Toeplitz operators. Until now researchers have been unable to characterize subnormalToeplitz operators in terms of their symbols. Thus the following question is very interesting andchallenging:

Which Toeplitz operators are subnormal ? (4.2)


The most interesting partial answer to Halmos’s Problem 5 was given by M. Abrahamse [Ab].M. Abrahamse gave a general sufficient condition for the answer to Halmos’s Problem 5 to beaffirmative. Abrahamse’s Theorem can be then stated as follows: Let ϕ = g+ f ∈ L∞ (f, g ∈ H2)be such that ϕ or ϕ is of bounded type. If Tϕ is subnormal then Tϕ is normal or analytic. In fact,it was also shown (cf. [CuL2], [CuL3]) that every 2-hyponormal Toeplitz operator with a boundedtype symbol is normal or analytic, and hence subnormal. On the other hand, very recently, theauthors of [CHL] have extended Abrahamse’s Theorem to block Toeplitz operators.

Theorem 4.1. (Extension of Abrahamse’s Theorem) (Curto-Hwang-Lee [CHL]) Suppose Φ = Φ∗−+Φ+ ∈ L∞Mn

is such that Φ and Φ∗ are of bounded type of the form

Φ− = B∗Θ (B ∈ H2Mn

; Θ = Iθ with an inner function θ) ,

where B and Θ are coprime. If TΦ is hyponormal and ker [T ∗Φ, TΦ] is invariant under TΦ then TΦ

is normal or analytic. Hence, in particular, if TΦ is subnormal then TΦ is normal or analytic.

We note that if n = 1 (i.e., TΦ is a scalar-valued Toeplitz operator), then Φ− = bθ withb ∈ H2. Thus, it automatically holds that b and θ are coprime. Consequently, if n = 1 thenTheorem 4.1 reduces to Abrahamse’s Theorem.

On the other hand, the study of square-hyponormality originated in [Hal3, Problem 209]. Itis easy to see that every power of a normal operator is normal and the same statement is true forevery subnormal operator. How about hyponormal operators? [Hal3, Problem 209] shows thatthere exists a hyponormal operator whose square is not hyponormal (e.g., U∗+2U for the unilateralshift U). However, as we remarked in the preceding, there exist special classes of operators forwhich square-hyponormality and subnormality coincide. For those classes of operators, it sufficesto check the square-hyponormality to show subnormality. This certainly gives a nice answer toquestion (4.2). Indeed, in [CuL1], it was shown that every hyponormal trigonometric Toeplitzoperator whose square is hyponormal must be either normal or analytic, and hence subnormal. In[Gu3], C. Gu showed that this result still holds for Toeplitz operators Tϕ with rational symbols ϕ(more generally, the cases where both ϕ and ϕ are of bounded type).

The aim of this section is to prove that this result can be extended to the block Toeplitzoperators whose symbols are matrix-valued rational functions.

We begin with:

Lemma 4.2. Suppose Φ = Φ∗− + Φ+ ∈ L∞Mnis a matrix-valued rational function of the form

Φ− = B∗Θ (coprime factorization) and Φ+ = ΘΘ0A∗,

where Θ = Iθ and Θ0 = Iθ0 with finite Blaschke products θ, θ0 and A,B ∈ H2Mn

. If TΦ ishyponormal then A(α) is invertible for each α ∈ Z(θ) \ Z(θ0).

Proof. Assume to the contrary that A(α) is not invertible for some α ∈ Z(θ) \ Z(θ0). Then byLemma 2.3, A and Bα := Ibα are not right coprime. Thus there exists a nonconstant inner matrixfunction ∆ such that

Bα = ∆1∆ = ∆∆1 and A = A1∆.

Write Θ := BαΘ′ = Θ′Bα. Then we may write Φ+ = Θ0Θ′∆1A∗1. Since TΦ is hyponormal, it

follows that

Θ0Θ′∆1H2Cn ⊆ kerHΦ∗+

⊆ kerHΦ∗−= ΘH2

Cn ,

which implies that Θ is a (left) inner divisor of Θ0Θ′∆1 (cf. [FF, Corollary IX.2.2]). Observe that

Θ is a (left) inner divisor of Θ0Θ′∆1 =⇒ Θ∗Θ0Θ′∆1 ∈ H2Mn

=⇒ Θ0∆1Θ′Θ∗ ∈ H2Mn

=⇒ Θ0∆∗ ∈ H2Mn

,


which implies that ∆ is a (right) inner divisor of Θ0. But since Bα and Θ0 are coprime, it followsthat ∆ and Θ0 are coprime. Thus ∆ is a constant unitary, a contradiction. This completes theproof.

Lemma 4.3. Suppose F,Bλ ∈ H∞Mn(Bλ := Ibλ). If G = GCD`F,Bλ, then G is a Blaschke-

Potapov factor of the form G = bλPN + (I − PN ) with

N :=(ranF (λ)

)⊥.

Proof. By assumption, G = GCDrF , Bλ. Then by Lemma 2.5,

G = bλPN + (I − PN ) for a closed subspace N .

Thus F = LG for some L ∈ H2Mn

, where L and BλG∗ = PN + bλ(I − PN ) are right coprime. We

argue that ker L(λ) ∩ ran (I − PN ) = 0. Indeed, if ker L(λ) ∩ ran (I − PN ) =: N0 6= 0 then

PN⊥0 + bλ(I − PN⊥0 ) would be a right inner divisor of L and PN + bλ(I − PN ) as follows:

L(λ) =

[∗ 0∗ 0

]N⊥0N0

=

[∗ 0∗ 0

] [1 00 bλ

]N⊥0N0

,

and hence,

L = L(λ) + CBλ = D

[1 00 bλ

]N⊥0N0

(some C,D ∈ H2Mn

)

and

PN + bλ(I − PN ) =

1 0 00 bλ 00 0 bλ

NN ′N0

=

1 0 00 bλ 00 0 1

1 0 00 1 00 0 bλ

NN ′N0

,

where N ′ := N⊥ N0. But since F (λ) = L(λ)(bλPN + (I − PN )

)(λ) = L(λ)(I − PN ), it follows

that

N = ker (I − PN ) = ker F (λ) = kerF ∗(λ) =(ranF (λ)

)⊥.

Therefore G = bλPN + (I − PN ) with N :=(ranF (λ)

)⊥.

Lemma 4.4. Let Φ ≡ Φ+ ∈ H∞Mnbe a matrix-valued rational function of the form

Φ = Θ∆rA∗r , (right coprime factorization),

= A∗`Ω, (left coprime factorization),

where Θ = Iθ with a finite Blaschke product θ and ∆r,Ω are inner matrix functions. Then Θ isan inner divisor of Ω.

Proof. Since ∆r is a finite Blaschke-Potapov product, we may write

∆r = ν

M∏m=1

(bmPm + (I − Pm)

)(bm :=

z − αm1− αmz

).

Without loss of generality we may assume that ν = In. Define

θ0 := GCDω : ω is inner, ∆r is an inner divisor of Ω = ωIn

.


Then θ0 =∏Mm=1 bm. Observe that

Φ = Θ∆rA∗r

= Θ

M∏m=1

(bmPm + (I − Pm)

)A∗r

=

M−1∏m=1

(bmPm + (I − Pm)

)BM

(PM + bM (I − PM )

)∗A∗rΘ (Bm := Ibm)

=

M−1∏m=1

(bmPm + (I − Pm)

)[Ar

(PM + bM (I − PM )

)]∗BMΘ

If PM = I, then

Φ =

M−1∏m=1

(bmPm + (I − Pm)

)A∗rBMΘ,

where Θ and Ar are coprime. If instead PM 6= I, then there are two cases to consider.

Case 1: Let αM /∈ Z(θ). Then

Φ =

M−1∏m=1

(bmPm + (I − Pm)

)A∗1BMΘ (with A1 := Ar(PM + bM (I − PM ))),

where Θ and A1 are coprime (by passing to Lemma 2.3).

Case 2: Let αM ∈ Z(θ). Write ΩM := GCD`BM , Ar(PM + bM (I − PM )

). Then we can

write

BM = ΩMΩ′M and Ar

(PM + bM (I − PM )

)= ΩMΓM (4.3)

for some Ω′M ,ΓM ∈ H∞Mn. By Lemma 4.3, ΩM = bMPN+(I−PN ) with N :=

(ran (Ar(αM )PM )

)⊥.

We now claim that

ΓM (αM ) is invertible. (4.4)

Since

det[Ar

(PM + bM (I − PM )

)]= b

rank(I−PM )M · detAr

and

det ΩMΓM = (bM )dimN · det ΓM ,

it follows from (4.3) that

detAr · (bM )rank(I−PM ) = (bM )dimN · det ΓM . (4.5)

But since Ar and Θ are right coprime, and hence Ar(αM ) is invertible, it follows that

dimN = dim(ran (Ar(αM )PM )

)⊥= dim(ranPM )⊥ = rank(I − PM ),

which together with (4.5) implies that det ΓM = detAr. This proves (4.4). Therefore we have

Φ =

M−1∏m=1

(bmPm + (I − Pm)

)Γ∗MΩ′MΘ,

where ΓM (αM ) is invertible. Thus ΓM (α) is invertible for all α ∈ Z(θ), and hence by Lemma 2.3,Θ and ΓM are coprime.

If we repeat this argument then afterM steps we get the left coprime factorization of Φ = A∗l Ω,where Ω still has Θ as an inner divisor.

Our main theorem of this section now follows:


Theorem 4.5. Let Φ ∈ L∞Mnbe a matrix-valued rational function. Then we may write

Φ− = B∗Θ ,

where B ∈ H2Mn

and Θ := Iθ with a finite Blaschke product θ. Suppose B and Θ are coprime. If

both TΦ and T 2Φ are hyponormal then TΦ is either normal or analytic.

Proof. Suppose Φ is not analytic. Then Θ is not constant unitary. Since TΦ is hyponormal, itfollows that kerHΦ∗+

⊆ kerHΦ∗−= ΘH2

Cn . Thus we can write

Φ+ = Θ∆rA∗r (right coprime factorization),

where ∆r is an inner matrix function. Let θ0 be a minimal inner function such that Θ0 ≡ Iθ0 =∆rΘ1 for some inner matrix function Θ1. We also write A := ArΘ1, and hence

Φ+ = ΘΘ0A∗.

On the other hand, we need to keep in mind that Θ = Iθ and Θ0 = Iθ0 are inner functions,constant along the diagonal, so that these factors commute with all other matrix functions in thecomputations below. Note that Φ2Θ2Θ2

0 ∈ H∞Mnand Φ∗2Θ2Θ2

0 ∈ H∞Mn. We thus have

T ∗Θ2Θ20[T 2∗

Φ , T 2Φ]TΘ2Θ2

0= T ∗Θ2Θ2

0T 2∗

Φ T 2ΦTΘ2Θ2

0− T ∗Θ2Θ2

0T 2

ΦT2∗Φ TΘ2Θ2

0

= TΦ∗2Θ∗2Θ∗20TΦ2Θ2Θ2

0− TΦ2Θ∗2Θ∗20

TΦ∗2Θ2Θ20

= TΦ∗2Θ∗2Θ∗20 Φ2Θ2Θ20− TΦ2Θ∗2Θ∗20 Φ∗2Θ2Θ2

0

= TΦ∗2Φ2 − TΦ2Φ∗2 = 0 (since Φ is normal).

The positivity of [T 2∗Φ , T 2

Φ] implies that [T 2∗Φ , T 2

Φ]TΘ2Θ20

= 0. We thus have

0 = [T 2∗Φ , T 2

Φ]TΘ2Θ20

= T 2Φ∗TΦ2Θ2Θ2

0− T 2

ΦTΦ∗2Θ2Θ20

= TΦ∗TΦ∗Φ2Θ2Θ20− TΦTΦΦ∗2Θ2Θ2

0

=(TΦ∗2Φ2Θ2Θ2

0−H∗ΦHΦ∗Φ2Θ2Θ2

0

)−(TΦ2Φ∗2Θ2Θ2

0−H∗Φ∗HΦΦ∗2Θ2Θ2

0

)(by (1.3))

= H∗Φ∗HΦΦ∗2Θ2Θ20−H∗ΦHΦ∗Φ2Θ2Θ2

0

= H∗Φ∗+HΦΦ∗2Θ2Θ20−H∗Φ∗−HΦ∗Φ2Θ2Θ2

0.

(4.6)

Let Ω := GCD (Θ0,Θ). Then by Lemma 2.1, Ω = Iω for an inner function ω. Thus we can write

Θ = Θ′Ω and Θ0 = Θ′0Ω,

where Θ′ = Iθ′ and Θ′0 = Iθ′0 for some inner functions θ′ and θ′0. Observe that

HΦ∗Φ2Θ2Θ20

= H(ΘB∗+Θ∗Θ∗0A)(Θ∗B+ΘΘ0A∗)2Θ2Θ20

= H(ΘB∗+Θ∗Θ∗0A)(B+Θ2Θ0A)2Θ20

= HΘ∗Θ∗0A(B+Θ2Θ0A)2Θ20

= HA(B+Θ2Θ0A)2Θ0Θ∗

= HA(B+Θ2Θ0A)2Θ′0Θ′∗ .

(4.7)

Since Φ is normal we also have

HΦΦ∗2Θ2Θ20

= H(ΘB∗+Θ∗Θ∗0A)2(Θ∗B+ΘΘ0A∗)Θ2Θ20

= H(Θ2Θ0B∗+A)(Θ∗B+ΘΘ0A∗)

= H(Θ2Θ0B∗+A)BΘ∗

= HABΘ∗ .

(4.8)

We now claim that

θ′ is not constant. (4.9)


Toward (4.9) we assume to the contrary that θ′ is a constant. Then by (4.7) we have

HΦ∗Φ2Θ2Θ20

= HA(B+Θ2Θ0A)2Θ′0Θ′∗ = 0 ,

so that H∗Φ∗−HΦ∗Φ2Θ2Θ2

0= 0, and by (4.6) we have

H∗Φ∗+HΦΦ∗2Θ2Θ20

= 0. (4.10)

Observe thatHABΘ∗ = 0⇐⇒ ABΘ∗ ∈ H2

Mn

⇐⇒ AB ∈ ΘH2Mn

⇐⇒ A = ΘA′ (since B and Θ are coprime) ,

which implies that A(α) = 0 for each α ∈ Z(θ), a contradiction. Therefore

HΦΦ∗2Θ2Θ20

= HABΘ∗ 6= 0 and cl ranHΦΦ∗2Θ2Θ20

= H(∆)

for some nonconstant (left) inner divisor ∆ of Θ. Thus it follows from Lemma 4.4 and (4.10) that

H(∆) = cl ranHΦΦ∗2Θ2Θ20⊆ kerH∗Φ∗+ ⊆ ΘH2

Cn ,

giving a contradiction. This proves (4.9). Observe

cl ranHΦ∗Φ2Θ2Θ20

= cl ranHA(B+Θ2Θ0A)2Θ′0Θ′∗ ⊆ H(Θ′) ⊥ ΘH2Cn = kerH∗Φ∗− ,

and

cl ranHΦΦ∗2Θ2Θ20

= cl ranHABΘ∗ ⊆ H(Θ) ⊥ ΘH2Cn ⊇ kerH∗Φ∗+ .

Thus by (4.6) we havekerHΦ∗Φ2Θ2Θ2

0= kerH∗Φ∗−HΦ∗Φ2Θ2Θ2

0

= kerH∗Φ∗+HΦΦ∗2Θ2Θ20

= kerHΦΦ∗2Θ2Θ20.

(4.11)

Observe that for all α ∈ Z(θ),(A(B + Θ2Θ0A)2Θ′0

)(α) = A(α)B(α)2Θ′0(α) .

Since B(α) and Θ′0(α) are invertible, we have

dim ker(A(B + Θ2Θ0A)2Θ′0

)(α) = dim kerA(α) = dim ker (AB)(α).

By (4.7), (4.8) and (4.11), we have that A(α) = 0 for all α ∈ Z(ω) and hence ω is a constant.Thus Ω is a constant unitary, and hence Θ = Θ′ and Θ0 = Θ′0. Therefore Z(θ) = Z(θ) \ Z(θ0)and hence, by Lemma 4.2, A(α) is invertible for each α ∈ Z(θ). Since for each α ∈ Z(θ),(

A(B + Θ2Θ0A)2Θ0

)(α) = A(α)B(α)2Θ0(α) and (AB)(α) are invertible,

it follows that A(B+ Θ2Θ0A)2Θ0 and Θ are coprime, and AB and Θ are coprime. Thus by (4.7),(4.8), and (4.11) we have

cl ranHΦ∗Φ2Θ2Θ20

= cl ranHΦΦ∗2Θ2Θ20

= H(Θ). (4.12)

By the well-known result of C. Cowen [Co1, Theorem 1] - if ϕ ∈ L∞ and b is a finite Blaschkeproduct of degree n then Tϕb ∼= ⊕nTϕ, we may, without loss of generality, assume that 0 ∈ Z(θ).Since TΦ is hyponormal, by [GHR] (cf. p.4) there exists K ∈ H∞Mn

with ||K||∞ ≤ 1 such that

HΦ∗−= HKΦ∗+

= T ∗KHΦ∗+

.

Since ΦΦ∗Θ2Θ20 ∈ H∞Mn

, we have

TKHΦ2Φ∗Θ2Θ20

= TKHΦTΦΦ∗Θ2Θ20

= TKHΦ∗−TΦΦ∗Θ2Θ2

0

= TK(T ∗KHΦ∗+

)TΦΦ∗Θ2Θ20

= TKT∗KHΦΦ∗2Θ2Θ2

0.


Thus by (4.6) we have0 = H∗Φ∗+HΦΦ∗2Θ2Θ2

0−H∗Φ∗−HΦ∗Φ2Θ2Θ2

0

= H∗Φ∗+

(HΦΦ∗2Θ2Θ2

0− TKHΦ∗Φ2Θ2Θ2

0

)= H∗Φ∗+(I − TKT

∗K

)HΦΦ∗2Θ2Θ20.

(4.13)

It thus follows from (4.12), (4.13) and Lemma 4.4 that

(I − TKT∗K

)(H(Θ)

)= cl ran

((I − TKT

∗K

)HΦ∗2ΦΘ2Θ20

)⊆ kerH∗Φ∗+ ⊆ ΘH2

Cn . (4.14)

Since ||K||∞ = ||K||∞ = ||K∗||∞, it follows that ||TK || = ||TK∗ || ≤ 1. For each i = 1, 2, · · · , n,put

Ei := (0, 0, · · · , 1, 0, · · · , 0, 0)t.

Since 0 ∈ Z(θ) ∩ Z(θ), we have Ei ∈ H(Θ) ∩H(Θ) and by (4.14),

Ei − TKT∗KEi = ΘFi (some Fi ∈ H2

Cn).

Observe that

Ei − TKT∗KEi = ΘFi =⇒ TKT

∗KEi = Ei − ΘFi

=⇒ ||TKT∗KEi||22 = ||Ei − ΘFi||22

=⇒ ||TKT∗KEi||22 = ||Ei||22 + ||ΘFi||22 (since Ei ∈ H(Θ))

=⇒ ||TKT∗KEi||22 = 1 + ||ΘFi||22.

But since ||TKT∗K|| ≤ 1, it follows that ||TKT

∗KEi||2 = 1 and Fi = 0 for all i = 1, 2, · · · , n. We

thus have ||T ∗KEi||2 = 1 for all i = 1, 2, · · · , n. Write

K(z) :=[kij(z)

]and kij(z) =

∞∑m=0

k(m)ij zm.

Then K∗(z) = K(z) =[kij(z)

], and hence

T ∗KEi = [P (k1i(z)), P (k2i(z)), · · · , P (kni(z))]

t = [k(0)1i , k

(0)2i , · · · , k

(0)ni ]t.

But since ||T ∗KEi||2 = 1 for all i = 1, 2, · · · , n, it follows that

||[k(0)1i , k

(0)2i , · · · , k

(0)ni ]t||2 = 1 for all i = 1, 2, · · ·n.

Therefore

1 =1

n

∣∣∣∣∣∣[k(0)ij

]∣∣∣∣∣∣22≤ 1

n

∞∑m=0

∣∣∣∣∣∣[k(m)ij

]∣∣∣∣∣∣22=

1

n||K||22 ≤ ||K||2∞ ≤ 1,

which implies that[k

(m)ij

]= 0 for all m ≥ 1. Hence K =

[k

(0)ij

], so that K = K∗. Observe that

(I − TKT∗K

)H(Θ) = 0 =⇒ (I − TK∗K)H(Θ) = 0

=⇒ K∗K = In (since 0 ∈ Z(θ)).

Therefore K = K∗ is a constant unitary and hence we have

[T ∗Φ, TΦ] = H∗Φ∗HΦ∗ −H∗ΦHΦ = H∗Φ∗HΦ∗ −H∗Φ∗TKK∗HΦ∗ = 0,

which implies that TΦ is normal.

Corollary 4.6. Let Φ ∈ L∞Mnbe a matrix-valued trigonometric polynomial whose co-analytic outer

coefficient is invertible. If TΦ and T 2Φ are hyponormal then TΦ is normal.

Proof. Immediate from Theorem 4.5 together with the observation that Φ− = B∗Θ with Θ = Izm

is a coprime factorization if and only if B(0) is a co-analytic outer coefficient and is invertible.


Remark 4.7. In Theorem 4.5, the “coprime” condition is essential. To see this, let

TΦ :=

[Tb + T ∗b 0

0 Tb

](b is a finite Blaschke product).

Since Tb + T ∗b is normal and Tb is analytic, it follows that TΦ and T 2Φ are both hyponormal.

Obviously, TΦ is neither normal nor analytic. Note that Φ− ≡ [ b 00 0 ] = [ 1 0

0 0 ]∗ · Ib, where [ 1 0

0 0 ] andIb are not coprime.

On the other hand, we have not been able to determine whether this phenomenon is quiteaccidental. In fact we would guess that if Φ ∈ L∞Mn

is a matrix-valued rational function such thatTΦ is subnormal then TΦ = TA ⊕ TB , where TA is normal and TB is analytic.

5. Subnormal Toeplitz Completions

Given a partially specified operator matrix with some known entries, the problem of finding suitableoperators to complete the given partial operator matrix so that the resulting matrix satisfies certaingiven properties is called a completion problem. Dilation problems are special cases of completionproblems: in other words, the dilation of T is a completion of the partial operator matrix

[T ?? ?

].

In recent years, operator theorists have been interested in the subnormal completion problem for[U∗ ?? U∗

], where U is the shift on H2. In this section, we solve this completion problem.

A partial block Toeplitz matrix is simply an n× n matrix some of whose entries are specifiedToeplitz operators and whose remaining entries are unspecified. A subnormal completion of a par-tial operator matrix is a particular specification of the unspecified entries resulting in a subnormaloperator. For example [

Tz 1− TzTz0 Tz

](5.1)

is a subnormal (even unitary) completion of the 2× 2 partial operator matrix[Tz ?? Tz

].

A subnormal Toeplitz completion of a partial block Toeplitz matrix is a subnormal completionwhose unspecified entries are Toeplitz operators. Then the following question comes up at once:Does there exist a subnormal Toeplitz completion of

[Tz ?? Tz

]? Evidently, (5.1) is not such a

completion. To answer this question, let

Φ ≡[z ϕψ z

](ϕ,ψ ∈ L∞).

If TΦ is hyponormal then by [GHR] (cf. p.4), Φ should be normal. Thus a straightforward calcu-lation shows that

|ϕ| = |ψ| and z(ϕ+ ψ) = z(ϕ+ ψ),

which implies that ϕ = −ψ. Thus a direct calculation shows that

[T ∗Φ, TΦ] =

[∗ ∗∗ TzTz − 1

],

which is not positive semi-definite because TzTz − 1 is not. Therefore, there are no hyponormalToeplitz completions of

[Tz ?? Tz

]. However the following problem has remained unsolved until now:

Problem A. Let U be the shift on H2. Complete the unspecified Toeplitz entries of the partialblock Toeplitz matrix A :=

[U∗ ?? U∗

]to make A subnormal.

In this section we give a complete answer to Problem A.


Theorem 5.1. Let ϕ,ψ ∈ L∞ and consider

A :=

[Tz TϕTψ Tz

].

The following statements are equivalent.

(i) A is normal.

(ii) A is subnormal.

(iii) A is 2-hyponormal.

(iv) One of the following conditions holds:

1. ϕ = eiθz + β and ψ = eiωϕ (β ∈ C; θ, ω ∈ [0, 2π));

2. ϕ = α z + eiθ√

1 + |α|2 z + β and ψ = ei (π−2 argα)ϕ

(α, β ∈ C, α 6= 0; θ ∈ [0, 2π)).

Theorem 5.1 says that the unspecified entries of the matrix[Tz ?? Tz

]are Toeplitz operators

with symbols which are both analytic or trigonometric polynomials of degree 1. In fact, as we willsee in the proof of Theorem 5.1, our solution is just the normal completion. However the solutionis somewhat more intricate than one would expect.

To prove Theorem 5.1 we need several technical lemmas.

Lemma 5.2. For j = 1, 2, 3, let θj be an inner function. If θ1H(θ2) ⊆ H(θ3) then either θ2 isconstant or θ1θ2 is a divisor of θ3. In particular, if θ1H(θ2) ⊆ H(θ1) then θ1 or θ2 is constant.

Proof. Suppose θ2 is not constant. If θ1H(θ2) ⊆ H(θ3) then by Lemma 2.6, for all f ∈ H(θ2),θ1fθ3 ∈ zH2, and hence fθ3 ∈ zH2, so that f ∈ H(θ3), which implies that H(θ2) ⊆ H(θ3), andtherefore θ3H

2 ⊆ θ2H2. Thus θ2 is a divisor of θ3. We can then write θ3 = θ0θ2 for some inner

function θ0. It suffices to show that θ1 is a divisor of θ0. Observe that

θ1H(θ2) ⊆ H(θ0θ2) =⇒ ran (Tθ1H∗θ2

) ⊆ H(θ0θ2)

=⇒ θ0θ2H2 ⊆ kerHθ2

Tθ1

=⇒ Hθ2Tθ1θ0θ2

= 0

=⇒ Hθ1θ0− Tθ2Hθ1θ0θ2

= 0

=⇒ Hθ1θ0− Tθ2Tθ2

Hθ1θ0= 0

=⇒ Hθ2Hθ0θ1

= 0 ,

where the fourth implication follows from the fact that Hϕψ = T ∗ϕHψ +HϕTψ for any ϕ,ψ ∈ L∞.But since θ2 is not constant it follows that θ1 is a divisor of θ0. The second assertion follows atonce from the first.

Suppose Φ ≡ Φ∗− + Φ+ ∈ L∞Mnis such that Φ and Φ∗ are of bounded type, with

Φ+ = A∗Θ and Φ− = B∗`Ω2 (left coprime factorization),

where Θ = Iθ for an inner function θ. If TΦ is hyponormal, then in view of Proposition 3.2, Φ canbe written as:

Φ+ = A∗Ω1Ω2 and Φ− = B∗`Ω2 , (5.2)

where Ω1Ω2 = Θ = Iθ. We also note that Ω1Ω2 = Θ = Ω2Ω1.

The following lemma will be extensively used in the proof of Theorem 5.1.


Lemma 5.3. Let Φ ≡ Φ∗−+ Φ+ ∈ L∞Mnbe such that Φ and Φ∗ are of bounded type of the form (5.2):

Φ+ = A∗Ω1Ω2 = A∗Θ and Φ− = B∗`Ω2 (left coprime factorization),

where Θ = Iθ for an inner function θ. If ker [T ∗Φ, TΦ] is invariant under TΦ, then

Ω1H2Cn ⊆ ker [T ∗Φ, TΦ],

and therefore

cl ran [T ∗Φ, TΦ] ⊆ H(Ω1).

Assume instead that we decompose Φ ∈ L∞Mnas:

Φ+ = ∆2∆0A∗r (right coprime factorization)

and

Φ− = ∆2B∗r (right coprime factorization) .

If TΦ is hyponormal then

∆2H(∆0) ⊆ cl ran [T ∗Φ, TΦ].

Hence, in particular, if TΦ is hyponormal and ker [T ∗Φ, TΦ] is invariant under TΦ, then

∆2H(∆0) ⊆ cl ran [T ∗Φ, TΦ] ⊆ H(Ω1). (5.3)

Proof. See [CHL, Lemma 3.2 and Theorem 3.7].

Lemma 5.4. Let

Φ− =

[z θ1bθ0a z

](a ∈ H(θ0), b ∈ H(θ1) and θj inner (j = 0, 1)).

If θ0 = znθ′0 (n ≥ 1; θ′0(0) 6= 0) and θ1(0) 6= 0, then kerHΦ∗−= ∆H2

C2 , where

∆ =

[zθ1 0

0 θ0

](n = 1);

1√|α|2+1

[zθ1 αθ1

−αθ0 zn−1θ′0

](n ≥ 2)

(α := − a(0)

θ1(0)

).

Proof. Observe that for f, g ∈ H2,

Φ∗−

[fg

]=

[z znθ′0a

θ1b z

] [fg

]∈ H2

C2 ⇐⇒[zf + znθ′0ag

θ1bf + zg

]∈ H2

C2 .

Thus if Φ∗−[fg

]∈ H2

C2 , then θ1bf + zg ∈ H2. Since θ1(0) 6= 0, we have θ1bfz ∈ H2, and hence

f = θ1f1 for some f1 ∈ H2. In turn, bf1 + zg ∈ H2, so that g = zg1 for some g1 ∈ H2. Wetherefore have

zθ1f1 + zn−1θ′0ag1 ∈ H2. (5.4)

If n = 1, then (5.4) implies g1 = θ′0g2 and f1 = zf2 for some g2, f2 ∈ H2. Thus f = zθ1f2 andg = θ0g2, which implies

kerHΦ∗−=

[zθ1 00 θ0

]H2

C2 .

If instead n ≥ 2, then (5.4) implies that zn−2θ′0ag1 ∈ H2, so that g1 = zn−2θ′0g2 for some g2 ∈ H2.We thus have

zθ1f1 + zn−1θ′0ag1 ∈ H2 ⇐⇒ zθ1f1 + zag2 ∈ H2

⇐⇒ θ1(0)f1(0) + a(0)g2(0) = 0

⇐⇒ g2(0) =1

αf1(0) (recall that α = − a(0)

θ1(0)).


Therefore we have[fg

]∈ kerHΦ∗−

⇐⇒ f = θ1f1, g = zn−1θ′0g2, and g2(0) =1

αf1(0). (5.5)

Put

∆ :=1√|α|2 + 1

[zθ1 αθ1

−αθ0 zn−1θ′0

].

Then ∆ is inner, and for h1, h2 ∈ H2,

∆

[h1

h2

]=

1√|α|2 + 1

[zθ1h1 + αθ1h2

−αznθ′0h1 + zn−1θ′0h2

]=

1√|α|2 + 1

[θ1

(zh1 + αh2

)zn−1θ′0

(−αzh1 + h2

)] .But since 1

α

(zh1 + αh2

)(0) =

(−αzh1 + h2

)(0), it follows from (5.5) that kerHΦ∗−

= ∆H2C2 .

Lemma 5.5. Let

Φ− =

[z θ1bθ0a z


If θ1 = znθ′1 (n ≥ 1; θ′1(0) 6= 0) and θ0(0) 6= 0, then kerHΦ∗−= ∆H2

C2 , where

∆ =

[θ1 0

0 zθ0

](n = 1);

1√|α|2+1

[zn−1θ′1 −αθ1

αθ0 zθ0

](n ≥ 2)

(α := − b(0)

θ0(0)

).

Proof. Same as the proof of Lemma 5.4.

Lemma 5.6. Let

Φ− =

[z θ1bθ0a z


If θ0 = zθ′0 and θ1 = zθ′1 then kerHΦ∗−= ∆H2

C2 , where

∆ =

[θ1 0

0 θ0

] ((ab)(0) 6= (θ′0θ

′1)(0)

);

1√|α|2+1

[θ1 αθ′1

−αθ0 θ′0

] ((ab)(0) = (θ′0θ

′1)(0)

) (α := − a(0)

θ′1(0)

).

Remark 5.7. Since a(0), b(0) 6= 0, the second part of the above assertion makes sense because byassumption, θ′0(0), θ′1(0) 6= 0.

Proof of Lemma 5.6. Observe that for f, g ∈ H2,

Φ∗−

[fg

]=

[z zθ′0a

zθ′1b z

] [fg

]∈ H2

C2 ⇐⇒[z(f + θ′0ag)

z(g + θ′1bf)

]∈ H2

C2

=⇒[f + θ′0ag

g + θ′1bf

]∈ zH2

C2

=⇒[θ′0ag

θ′1bf

]∈ H2

C2

=⇒ g = θ′0g1 and f = θ′1f1 for some g1, f1 ∈ H2.


Thus if Φ∗−[fg

]∈ H2

C2 then z(θ′1f1 + ag1) ∈ H2 and z(θ′0g1 + bf1) ∈ H2, so that

θ′1(0)f1(0) = −a(0)g1(0) and θ′0(0)g1(0) = −b(0)f1(0).

If (ab)(0) = (θ′0θ′1)(0), then

θ′1(0)f1(0) = −a(0)g1(0)⇐⇒ θ′0(0)g1(0) = −b(0)f1(0).

Put

∆ :=1√|α|2 + 1

[θ1 αθ′1

−αθ0 θ′0

].

Then we can see that ∆ is inner and kerHΦ∗−= ∆H2

C2 .

If (ab)(0) 6= (θ′0θ′1)(0), put θ0 = zmθ′′0 and θ1 = znθ′′1 (θ′′0 (0), θ′′1 (0) 6= 0). If n = m then

Φ− =

[z znθ′′1 b

zmθ′′0a z

]= Iznθ′′0 θ′′1

[zn−1θ′′0θ

′′1 θ′′0 b

θ′′1a zn−1θ′′0θ′′1

]≡ Iznθ′′0 θ′′1 B

∗.

Since B(0) is invertible, it follows from Lemma 2.3 that Izn and B are coprime. Observe that

Φ∗−

[znfzng

]=

[z znθ′′0a

znθ′′1 b z

] [znfzng

]∈ H2

C2 ⇐⇒[zn−1f + θ′′0ag

θ′′1 bf + zn−1g

]∈ H2

C2 ,

which implies

f = θ′′1f1 and g = θ′′0g1 for some f1, g1 ∈ H2.

We thus have

kerHΦ∗−=

[θ1 00 θ0

]H2

C2

If instead n 6= m, then

Φ∗−

[fg

]=

[z zmθ′′0a

znθ′′1 b z

] [fg

]=

[z(f + zm−1θ′′0ag)

z(g + zn−1θ′′1 bf)

]∈ H2

C2

⇐⇒

f + zm−1θ′′0ag ∈ zH2

g + zn−1θ′′1 bf ∈ zH2,

which implies

f = zn−1θ′′1f1 and g = zm−1θ′′0g1.

Suppose n > m, and hence n ≥ 2. We thus have

zn−2θ′′1f1 + zag1 ∈ H2 =⇒ zag1 ∈ H2 =⇒ g1 = zg2 =⇒ g = θ0g2.

In turn,

g + zn−1θ′′1 bf ∈ zH2 =⇒ zmθ′′0g2 + bf1 ∈ zH2 =⇒ f1 = zf2,

which implies

kerHΦ∗−=

[θ1 00 θ0

]H2

C2 .

If m > n, a similar argument gives the result.

We are ready for:

Proof of Theorem 5.1. Clearly (i) ⇒ (ii) and (ii) ⇒ (iii). Moreover, a simple calculation showsthat (iv) ⇒ (i).

(iii) ⇒ (iv): Write

Φ ≡[z ϕψ z

]≡ Φ∗− + Φ+ =

[z ψ−ϕ− z

]∗+

[0 ϕ+

ψ+ 0

]


and assume that TΦ is 2-hyponormal. Since ker [T ∗, T ] is invariant under T for every 2-hyponormaloperator T ∈ B(H), we note that Theorem 4.1 and Lemma 5.3 hold for 2-hyponormal operatorsTΦ. We claim that

|ϕ| = |ψ|, and (5.6)

Φ and Φ∗ are of bounded type. (5.7)

Indeed, if TΦ is hyponormal then Φ is normal, so that a straightforward calculation gives (5.6).

Also there exists a matrix function K ≡[k1 k2

k3 k4

]∈ H∞M2

with ||K||∞ ≤ 1 such that Φ−KΦ∗ ∈ H∞M2,

i.e., [z ϕ−ψ− z

]−[k1 k2

k3 k4

] [0 ψ+

ϕ+ 0

]∈ H2

M2,

which implies that Hz = Hk2ϕ+

= Hϕ+Tk2

;

Hϕ− = Hk1ψ+= Hψ+

Tk1;

Hψ−= Hk4ϕ+

= Hϕ+Tk4

;

Hz = Hk3ψ+= Hψ+

Tk3 .

If ϕ+ is not of bounded type then kerHϕ+= 0, so that k2 = 0, a contradiction; and if ψ+ is not

of bounded type then kerHψ+= 0, so that k3 = 0, a contradiction. Therefore we should have Φ∗

of bounded type. Since TΦ is hyponormal, Φ is also of bounded type, giving (5.7). Thus we canwrite

ϕ− := θ0a and ψ− := θ1b (a ∈ H(θ0), b ∈ H(θ1)).

Putθ0 = zmθ′0 and θ1 = znθ′1 (m,n ≥ 0; θ′0(0) 6= 0 6= θ′1(0)).

We now claim thatm = n = 0 or m = n = 1. (5.8)

We split the proof of (5.8) into three cases.

Case 1 (m 6= 0 and n = 0) In this case, we have a(0) 6= 0 because θ0(0) = 0 and θ0 and aare coprime. We first claim that m = 1. To show this we assume to the contrary that m ≥ 2.Write

α := − a(0)

θ1(0)and ν :=

1√|α|2 + 1

.

By Lemma 5.4, we can write:

Φ− =

[z θ1bθ0a z

]= ∆2B

∗r (right coprime factorization) ,

where

∆2 := ν

[zθ1 αθ1

−αθ0 zm−1θ′0

]and Br := ν

[θ1 − αa αθ1z + az

zb− αzm−1θ′0 αb+ zm−2θ′0

].

To get the left coprime factorization of Φ−, applying Lemma 5.4 for Φ− gives

Φ− =

[z θ0a

θ1a z

]= Ω2B

∗` (right coprime factorization) ,

where

Ω2 := ν

[zm−1θ′0 αθ1

−αθ0 zθ1

]and B` := ν

[zm−2θ′0 + αb az + αθ1z−αzm−1θ′0 + zb −αa+ θ1

],

which gives

Φ− =

[z θ1bθ0a z

]= B∗`Ω2 (left coprime factorization) .


On the other hand, since Φ∗− −KΦ∗+ ∈ H2Mn

, and hence[z θ0a

θ1b z

]−[k2ϕ+ k1ψ+

k4ϕ+ k3ψ+

]∈ H2

M2,

we have z − k2ϕ+ ∈ H2, θ1b− k4ϕ+ ∈ H2

z − k3ψ+ ∈ H2, θ0a− k1ψ+ ∈ H2,

which via Cowen’s Theorem gives that the following Toeplitz operators are all hyponormal:

Tz+ϕ+, Tθ1b+ϕ+

, Tz+ψ+, Tθ0a+ψ+

.

Thus by Proposition 3.2 we can see that

ϕ+ = zθ1θ3d and ψ+ = θ0θ2c for some inner functions θ2, θ3.

We thus have

Φ+ =

[zθ1θ3 0

0 θ0θ2

] [0 cd 0

]∗≡ ∆2∆0A

∗r (right coprime factorization),

where

Ar :=

[0 cd 0

];

∆0 := ν

[1 −ααz z

] [θ3 00 θ2

];

∆2 := ν

[zθ1 αθ1

−αθ0 zm−1θ′0

]=

[θ1 00 zm−1θ′0

]· ν[z α−αz 1

].

Write

θ2 = zpθ′2 and θ3 = zqθ′3 (p, q ≥ 0; θ′2(0), θ′3(0) 6= 0).

If q + 1 ≥ m+ p, then LCM (zθ1θ3, θ0θ2) is an inner divisor of zq+1θ′0θ1θ′2θ′3. Thus we can write

Φ+ =

[0 xy 0

]∗Izq+1θ′0θ1θ

′2θ′3≡ A∗Ω1Ω2,

where

A :=

[0 xy 0

](some x, y ∈ H2);

Ω1 :=

[zq+1−mθ1θ

′2θ′3 0

0 zqθ′0θ′2θ′3

]· ν[z −ααz 1

].

It thus follows from Lemma 5.3 that

∆2H(∆0) ⊆ cl ran [T ∗Φ, TΦ] ⊆ H(Ω1). (5.9)

But since in general, Θ2H(Θ1) ⊆ H(Θ1Θ2) for inner matrix functions Θ1,Θ2, we have

ν

[1 −ααz z

]H([θ3 0

0 θ2

])⊆ H(∆0).

Thus by (5.9), we have

∆2 · ν[

1 −ααz z

]H([θ3 0

0 θ2

])⊆ H(Ω1),

or equivalently, [zθ1 00 zmθ′0

]H([θ3 00 θ2

])⊆ H(Ω1). (5.10)

Since in general, F ∈ H(Θ) if and only if Θ∗F ∈ (H2Cn)⊥, (5.10) implies

ν

[1 α−αz z

]H([θ3 0

0 θ2

])⊆ H

([zq+1−mθ′2θ′3 0

0 zq+1−mθ′2θ′3

]). (5.11)


Also since for inner matrix functions Θ1,Θ2 and any closed subspace F of H2Cn ,

Θ1F ⊆ H(Θ1Θ2) and Θ1Θ2 = Θ2Θ1 =⇒ F ⊆ H(Θ1Θ2) ,

it follows from (5.11) that [H(θ3)H(θ2)

]⊆[H(zq+1−mθ′2θ

′3)

H(zq+1−mθ′2θ′3)

].

But since θ3 = zqθ′3, it follows that q + 1−m ≥ q, giving a contradiction.

If instead q + 1 < m+ p, then LCM (zθ1θ3, θ0θ2) is an inner divisor of zm+pθ′0θ1θ′2θ′3. Thus

we can write

Φ+ =

[0 xy 0

]∗Izm+pθ′0θ1θ

′2θ′3≡ A∗1Ω′1Ω2,

where

A1 :=

[0 xy 0

](some x, y ∈ H2);

Ω′1 :=

[zpθ1θ

′2θ′3 0

0 zm+p−1θ′0θ′2θ′3

]· ν[z −ααz 1

].

It thus follows from Lemma 5.3 with Ω′1 in place of Ω1 that

∆2H(∆0) ⊆ cl ran [T ∗Φ, TΦ] ⊆ H(Ω′1).

Since

H(∆0) = H(ν

[1 −ααz z

] [θ3 00 θ2

])= H

(ν

[1 −ααz z

]) ⊕ν

[1 −ααz z

]H([θ3 0

0 θ2

]),

we have

∆2H(ν

[1 −ααz z

])⊕[zθ1 00 zmθ′0

]H([θ3 0

0 θ2

])⊆ H(Ω′1). (5.12)

Then by the same argument as (5.10) and (5.11), we can see that

ν

[1 α−αz z

]H([θ3 00 θ2

])⊆ H

([zpθ′2θ

′3 0

0 zpθ′2θ′3

]), (5.13)

which gives

zH(θ2) ⊆ H(zpθ′2θ′3),

which by Lemma 5.2 implies that θ2 should be a constant. Thus (5.13) can be written as

ν

[1 α−αz z

]H([θ3 0

0 1

])⊆ H

([θ′3 00 θ′3

]),

which gives zH(θ3) ⊆ H(θ′3). It follows from again Lemma 5.2 that θ3 is a constant. Thus by(5.12), we have[

θ1 00 zm−1θ′0

]ν

[z α−αz 1

]H(ν

[1 −ααz z

])⊆ H

([θ1 00 zm−1θ′0

]ν

[z −ααz 1

]),

so that

ν

[z α−αz 1

]H(ν

[1 −ααz z

])⊆ H

(ν

[z −ααz 1

]),

giving a contradiction by Lemma 5.3. Therefore we should have

m = 1, i.e., θ0 = zθ′0.

Thus by Lemmas 5.5 and 5.6, we have

Φ− =

[z θ1b

zθ′0a z

]=

[zθ1 00 θ0

] [θ1 azb θ′0

]∗(right coprime factorization)


and

Φ− =

[z θ1b

zθ′0a z

]=

[θ′0 azb θ1

]∗ [θ0 00 zθ1

](left coprime factorization).

Recall that

ψ+ = θ0θ2c and ϕ+ = zθ1θ3d for some inner functions θ2 and θ3.

We can thus write

Φ+ =

[0 zθ1θ3d

θ0θ2c 0

]=

[zθ1θ3 0

0 θ0θ2

] [0 cd 0

]∗(right coprime factorization).

Note that LCM (zθ1θ3, θ0θ2) is an inner divisor of θ0θ1θ2θ3. Thus we can write

Φ+ =

[0 xy 0

]∗Iθ0θ1θ2θ3 (x, y ∈ H2).

It follows from Lemma 5.3 that[zθ1H(θ3)θ0H(θ2)

]⊆ cl ran [T ∗Φ, TΦ] ⊆

[H(θ1θ2θ3)H(θ′0θ2θ3)

],

which implies

zH(θ3) ⊆ H(θ2θ3) and zH(θ2) ⊆ H(θ2θ3).

By Lemma 5.2, either θ3 is constant or zθ3 is a divisor of θ2θ3;

either θ2 is constant or zθ2 is a divisor of θ2θ3 ,(5.14)

If θ2 or θ3 is not constant then it follows from (5.14) that z is a divisor of θ2 and θ3. Thus we havep, q ≥ 1. Let N := max(p, q). Then LCM (zθ1θ3, θ0θ2) is an inner divisor of zNθ0θ1θ

′2θ′3. Thus

we can write

Φ+ =

[0 xy 0

]∗IzNθ0θ1θ′2θ′3 (x, y ∈ H2).

It follows from Lemma 5.3 that[zθ1H(zqθ′3)θ0H(zpθ′2)


[H(zNθ1θ

′2θ′3)

H(zNθ′0θ′2θ′3)

]=⇒

zq is a divisor of zN−1θ′2zp is a divisor of zN−1θ′3,

giving a contradiction. Therefore

θ2 and θ3 are constant.

We observe that LCM(zθ1, θ0) is an inner divisor of zθ′0θ1. It follows from Lemma 5.3 that[θ1H

2

θ′0H2

]⊆ ker [T ∗Φ, TΦ].

In particular,[

0θ′0

]∈ ker [T ∗Φ, TΦ]. Observe that

Φ∗−

[0θ′0

]=

[z θ0a

θ1b z

] [0θ′0

]=

[zazθ′0

],

so that

HΦ∗−

[0θ′0

]=

[a(0)θ′0(0)

].

We thus have

H∗Φ∗−HΦ∗−

[0θ′0

]=

[θ′0(0)J(I − P )(θ1b) + a(0)

∗

].


A similar calculation shows that

H∗Φ∗+HΦ∗+

[0θ′0

]=

[0∗

].

Since [T ∗Φ, TΦ] = H∗Φ∗+HΦ∗+

−H∗Φ∗−HΦ∗−, it follows that

θ′0(0)J(I − P )(θ1b) = −a(0).

But since a(0) 6= 0, we must have that θ1b ∈ zH2 ∩ (H2)⊥, which implies that θ1 = c z for anonzero constant c, giving a contradiction because θ1(0) 6= 0. Therefore this case cannot occur.

Case 2 (m = 0 and n 6= 0) This case is symmetrical to Case 1. Thus the proof is identicalto that of Case 1. Therefore this case cannot occur either.

Case 3 (m 6= 0, n 6= 0 and m ≥ 2 or n ≥ 2) In this case we have a(0) 6= 0 and b(0) 6= 0because θ0(0) = θ1(0) = 0, θ0 and a are coprime, and θ1 and b are coprime. By Lemma 5.6 wehave

Φ− =

[θ1 00 θ0

] [zn−1θ′1 ab zm−1θ′0


Similarly, we have

Φ− =

[zm−1θ′0 a

b zn−1θ′1

]∗ [θ0 00 θ1

](left coprime factorization)

and

Φ+ =

[θ1θ3 0

0 θ0θ2

] [0 cd 0


We then claim that

θ2 and θ3 are constant. (5.15)

Assume θ2 is not constant. Put θ2 = zpθ′2 and θ3 = zqθ′3 (θ′2 6= 0 6= θ′3(0) and p, q ≥ 0) and letN := max (m+ p, n+ q). Then LCM(θ0θ2, θ1θ3) is a divisor of zNθ′0θ

′1θ′2θ′3. Thus we can write

Φ+ =

[0 xy 0

]∗IzNθ′0θ′1θ′2θ′3 (x, y ∈ H2).

By Lemma 5.3 we have [θ1H(θ3)θ0H(θ2)

]⊆[H(zN−mθ′1θ

′2θ′3)

H(zN−nθ′0θ′2θ′3)

].

If θ3 is a constant, then q = 0 and m+ p ≤ N − n, giving a contradiction because m,n ≥ 2. If θ3

is not constant, then n+ q ≤ N −m and m+ p ≤ N − n, giving a contradiction because m,n ≥ 2.Therefore we should have that θ2 is constant. Similarly, we can show that θ3 is also constant.This proves (5.15).

We now suppose n ≤ m. By Lemma 5.3 we have[θ′1H

2

zm−nθ′0H2

]⊆ ker [T ∗Φ, TΦ].

In particular,[θ′10


Φ∗−

[θ′10

]=

[z θ0a

θ1b z

] [θ′10

]=

[zθ′1znb

],

so that

HΦ∗−

[θ′10

]=

[θ′1(0)

zn−1b1

](b1 := PH(zn)(b)).


Put b3 := zn−1b1. We also have

Φ∗−

[θ′1(0)b3

]=

[z θ1b

θ0a z

] [θ′1(0)b3

]=

[∗

θ′1(0)θ0a+ b3z

],

so that


[θ′10

]=

[∗

θ′1(0)J(I − P )(θ0a) + b3(0)

].


H∗Φ∗+HΦ∗+

[θ′10

]=

[∗0

].

It thus follows that

θ′1(0)J(I − P )(θ0a) = −b3(0) =⇒ θ0a ∈ zH2 =⇒ θ0a ∈ zH2 =⇒ θ0a ∈ zH2 ∩(H2)⊥.

Since n ≤ m and (m ≥ 2 or n ≥ 2), it follows that m ≥ 2. Thus θ0 a ∈ zH2 ∩ (H2)⊥ implies

zm−1θ′0 a = c (a constant), which forces a = 0, giving a contradiction. If instead n > m then thesame argument leads a contradiction. Therefore this case cannot occur. This completes the proofof (5.8).

Now in view of (5.8) it suffices to consider the case m = n = 0 and the case m = n = 1.

Case A (m = n = 0). In this case, we first claim that

ϕ− = ψ− = 0, i.e., ϕ and ψ are analytic. (5.16)

Put θ := GCD(θ0, θ1). Then θ0 = θ′0θ and θ1 = θ′1θ for some inner functions θ′0, θ′1, and hence

LCM(θ0, θ1) = θθ′0θ′1. We thus have

Φ− = Izθθ′0θ′1

[θθ′0θ

′1 zθ′1a

zθ′0b θθ′0θ′1

]∗≡ Izθθ′0θ′1 B

∗ .

Since B(0) is invertible it follows from Lemma 2.3 that Iz and B are coprime. Observe that

Φ∗−

[zfzg

]=

[z θ0a

θ1b z

] [zfzg

]∈ H2

C2 ⇐⇒[θ0azg

θ1bzf

]∈ H2

C2

⇐⇒ g ∈ θ0H2, f ∈ θ1H

2,

which implies

kerHΦ∗−=

[zθ1 00 zθ0

]H2

C2 .

We thus have

Φ− =

[zθ1 00 zθ0

] [θ1 zazb θ0


To get the left coprime factorization of Φ−, we take

Φ− =

[zθ0 0

0 zθ1

][θ0 zb

za θ1

]∗(right coprime factorization),

which implies

Φ− =

[θ0 zazb θ1

]∗ [zθ0 00 zθ1

](left coprime factorization).

Since TΦ is hyponormal and hence,


=

[zθ1H

2

zθ0H2

],


it follows that

ψ+ = zθ0θ2c and ϕ+ = zθ1θ3d for some inner functions θ2, θ3.

We can thus write

Φ+ =

[0 zθ1θ3d

zθ0θ2c 0

]=

[zθ1θ3 0

0 zθ0θ2

] [0 cd 0


Observe that LCM (zθ1θ3, zθ0θ2) is an inner divisor of zθ0θ1θ2θ3. Thus we can write

Φ+ =

[0 xy 0

]∗Izθ0θ1θ2θ3 (x, y ∈ H2).

It follows from Lemma 5.3 that[zθ1H(θ3)zθ0H(θ2)


[H(θ1θ2θ3)H(θ0θ2θ3)

],

which implies thatzH(θ3) ⊆ H(θ2θ3) and zH(θ2) ⊆ H(θ2θ3).

Thus the same argument as in (5.14) shows that

θ2 and θ3 are constant.

We now observe that LCM(zθ1, zθ0) is an inner divisor of zθ0θ1. Thus we can write

Φ+ =

[0 xy 0

]∗Izθ0θ1 (x, y ∈ H2).

It follows from Lemma 5.3 that[θ1H

2

θ0H2

]⊆ ker [T ∗Φ, TΦ]. In particular,

[θ10

]∈ ker [T ∗Φ, TΦ]. Observe

that

Φ∗−

[θ1

0

]=

[z θ0a

θ1b z

] [θ1

0

]=

[zθ1

b

],

so that

HΦ∗−

[θ1

0

]=

[θ1(0)

0

].

We thus have


[θ1

0

]=

[θ1(0)

θ1(0)J(I − P )(θ0a)

].


H∗Φ∗+HΦ∗+

[θ1

0

]=

[∗0

].


θ1(0)J(I − P )(θ0a) = 0 =⇒ θ0a ∈ H2 =⇒ θ0a ∈ H2 =⇒ θ0a ∈ H2 ∩(H2)⊥,

which implies that a = 0 and hence φ is analytic. Similarly, we can show that ψ is also analytic.This gives (5.16).

Now since by (5.16), ϕ,ψ ∈ H∞ and |ϕ| = |ψ|, we can write ϕ = θ1a and ψ = θ2a, where theθi are inner functions and a is an outer function. Observe that

Φ− ≡ B∗Θ2,

where B ≡ [ 1 00 1 ] and Θ2 ≡ [ z 0

0 z ] are coprime. Thus our symbol satisfies all the assumptions ofTheorem 4.1. Thus by Theorem 4.1, since TΦ is 2-hyponormal then TΦ must be normal. We thushave

H∗Φ∗+HΦ∗+= H∗Φ∗−HΦ∗−

. (5.17)

Now observe that

Φ+ =

[0 ϕψ 0

]and Φ− =

[z 00 z

].


Since TΦ is normal we have [H∗ϕHϕ 0

0 H∗ψHψ

]=

[Hz 00 Hz

],

which impliesH∗ϕHϕ = Hz = H∗

ψHψ, (5.18)

which says that Hϕ and Hψ are both rank-one operators. Now remember that if T is a rank-

one Hankel operator then there exist ω ∈ D and a constant c such that T = c (kω ⊗ kω), wherekω := 1

1−ωz is the reproducing kernel for ω. Note that kω ⊗ kω is represented by the matrix1 ω ω2 ω3 . . .ω ω2 ω3 . . .ω2 ω3 . . .ω3

...

.By (5.18) we have that ω = 0. We thus have

ϕ = eiθ1z + β1 and ψ = eiθ2z + β2 (β1, β2 ∈ C, θ1, θ2 ∈ [0, 2π)). (5.19)

But since |ϕ| = |ψ|, we have

ϕ = eiθz + β and ψ = eiωϕ (β ∈ C, θ, ω ∈ [0, 2π)). (5.20)

Case B (m = n = 1) In this case, θ0 = zθ′0 and θ1 = zθ′1 (θ′0(0), θ′1(0) 6= 0). We thus have

ϕ− = zθ′0a and ψ− = zθ′1b,

so that

Φ− =

[z zθ′1b

zθ′0a z

].

There are two subcases to consider.

Case B-1 ((ab)(0) 6= (θ′0θ′1)(0)). In this case, we have, by Lemma 5.6,

Φ− =

[z zθ′1b

zθ′0a z

]=

[zθ′1 00 zθ′0

] [θ′1 ab θ′0

]∗(right coprime decompositon)

=

[θ′0 ab θ′1

]∗ [zθ′0 00 zθ′1

](left coprime factorization)

and

Φ+ =

[zθ′1θ3 0

0 zθ′0θ2

] [0 cd 0


Suppose θ2 is not constant. Put θ2 = zpθ′2 and θ3 = zqθ′3 (p, q ∈ N ∪ 0). Let N := max(p, q).Then LCM(θ1θ3, θ0θ2) is a divisor of zN+1θ′0θ

′1θ′2θ′3. Thus we can write

Φ+ =

[0 xy 0

]∗IzN+1θ′0θ

′1θ′2θ′3

(x, y ∈ H2).

By Lemma 5.3 we have [zθ′1H(θ3)zθ′0H(θ2)

]⊆[H(zNθ′1θ

′2θ′3)

H(zNθ′0θ′2θ′3)

].

If θ3 is a constant then p + 1 ≤ N = p, giving a contradiction. If instead θ3 is not constant thenq + 1 ≤ N and p + 1 ≤ N , giving a contradiction. The same argument gives θ3 is a constant.Therefore θ2 and θ3 should be constant. Note that LCM(zθ′1, zθ

′0) is an inner divisor of zθ′0θ

′1.

Thus we can write

Φ+ =

[0 xy 0

]∗Izθ′0θ′1 (x, y ∈ H2).


It follows from Lemma 5.3 that [θ′1H

2

θ′0H2

]⊆ ker [T ∗Φ, TΦ]. (5.21)

In particular,[θ′10


Φ∗−

[θ′10

]=

[z θ0a

θ1b z

] [θ′10

]=

[zθ′1zb

],

so that

HΦ∗−

[θ′10

]=

[θ′1(0)b(0)

].

We thus have

Φ∗−

[θ′1(0)b(0)

]=

[z θ1b

θ0a z

] [θ′1(0)b(0)

]=

[∗

θ′1(0)θ0a+ b(0)z

],

so that


[θ′10

]=

[∗

θ′1(0)J(I − P )(θ0a) + b(0)

].


H∗Φ∗+HΦ∗+

[θ′10

]=

[∗0

].


θ′1(0)J(I − P )(θ0a) = −b(0).

Since b(0) 6= 0, we have that θ0a ∈ zH2, which implies that θ0a = αz for a nonzero constant α.Therefore we must have that θ′0 is a constant. Similarly, we can show that θ1b = βz for a nonzeroconstant β, and hence θ′1 is also a constant. Therefore by (5.21), TΦ is normal. Now observe that

Φ+ =

[0 ϕ+

ψ+ 0

]and Φ∗− =

[z αzβz z

](α 6= 0 6= β).

Since TΦ is normal we have[H∗ϕ+

Hϕ+0

0 H∗ψ+Hψ+

]=

[(1 + |β|2)Hz (α+ β)Hz

(α+ β)Hz (1 + |α|2)Hz

],

which implies that β = −αH∗ϕ+

Hϕ+= (1 + |β|2)Hz

H∗ψ+Hψ+

= (1 + |α|2)Hz.

(5.22)

By the case assumption, 1 6= |ab| = |αβ| = |α|2, i.e., |α| 6= 1. By the same argument as in (5.18)we have

ϕ+ = eiθ1√

1 + |α|2 z + β1 and ψ+ = eiθ2√

1 + |α|2 z + β2,

(β1, β2 ∈ C; θ1, θ2 ∈ [0, 2π)) which implies that

ϕ = α z + eiθ1√

1 + |α|2 z + β1 and ψ = −α z + eiθ2√

1 + |α|2 z + β2.

Since |ϕ| = |ψ|, it follows that∣∣∣eiθ1√1 + |α|2 z2 + β1z + α∣∣∣ =

∣∣∣eiθ2√1 + |α|2 z2 + β2z − α∣∣∣ for all z on T.

We argue that if p and q are polynomials having the same degree and the outer coefficients of thesame modulus then

|p(z)| = |q(z)| on |z| = 1 =⇒ p(z) = eiωq(z) for some ω ∈ [0, 2π).


Indeed, if |p(z)| = |q(z)| on |z| = 1, then p = θq for a finite Blaschke product θ, i.e., p =∏nj=1

z−αj1−αjz q (|αj | ≤ 1). But since the modulus of the outer coefficients are same, it follows that∏n

j=1 |αj | = 1 and therefore, p = eiωq for some ω. Using this fact we can see that ψ = eiωϕ for

some ω ∈ [0, 2π). But then a straightforward calculation shows that ω = π − 2 argα, and hence

ϕ = α z + eiθ√

1 + |α|2 z + β and ψ = ei (π−2 argα)ϕ,

where α 6= 0, |α| 6= 1, β ∈ C, and θ ∈ [0, 2π).

Case B-2 ((ab)(0) = (θ′0θ′1)(0)). A similar argument as in Case 1 shows that θ2 and θ3 are

constant and the same argument as in Case B-1 gives that

ϕ = α z + eiθ√

1 + |α|2 z + β and ψ = ei (π−2 argα)ϕ,

where α 6= 0, |α| = 1, β ∈ C, and θ ∈ [0, 2π). We here note that the condition |α| = 1 comes fromthe case assumption 1 = |θ′0θ′1| = |ab| = |α|2.

Therefore if we combine the two subcases of Case B-1 and B-2 then we can conclude that

ϕ = α z + eiθ√

1 + |α|2 z + β and ψ = ei (π−2 argα)ϕ, (5.23)

where α 6= 0, β ∈ C, and θ ∈ [0, 2π). This completes the proof.

Remark 5.8. We would also ask whether there is a subnormal non-Toeplitz completion of[Tz ?? Tz

].

Unexpectedly, there is a normal non-Toeplitz completion of[Tz ?? Tz

]. To see this, let B be a

selfadjoint operator and put

T =

[Tz Tz +B

Tz +B Tz

].

Then

[T ∗, T ] =

[TzB +BTz − (TzB +BTz) TzB +BTz − (TzB +BTz)BTz + TzB − (BTz + TzB) TzB +BTz − (TzB +BTz)

],

so that T is normal if and only if

TzB +BTz = TzB +BTz, i.e., [Tz, B] = [Tz, B]. (5.24)

We define

α1 := 0 and αn := −2

3

(1−

(−1

2

)n)for n ≥ 2.

Let D ≡ diag (αn), i.e., a diagonal operator whose diagonal entries are αn (n = 1, 2, . . .) and foreach n = 1, 2 . . ., let Bn be defined by

Bn = − 1

2n−1diag(αn−1)T ∗z2n .

Then

||Bn|| ≤1

2n−1supαn−1 <

1

2n−1,

which implies that ∣∣∣∣∣∣∣∣∣∣∞∑n=1

Bn

∣∣∣∣∣∣∣∣∣∣ ≤ 2.

We define C by

C :=

∞∑n=1

Bn.


Then C looks like:

C =

0 0 1 0 12 0 1

22 0 · · ·0 0 0 1

2 0 122 0 1

23 · · ·0 0 0 0 3

22 0 323 0 · · ·

0 0 0 0 0 523 0 5

24 · · ·0 0 0 0 0 0 11

24 0 · · ·0 0 0 0 0 0 0 21

25 · · ·0 0 0 0 0 0 0 0 · · ·0 0 0 0 0 0 0 0 · · ·0 0 0 0 0 0 0 0 · · ·...

......

......

......

.... . .

.

Note that C is bounded. If we define B by

B := D + C + C∗,

then a straightforward calculation shows that B satisfies equation (5.24). Therefore the operator

T =

[Tz Tz +B

Tz +B Tz

]is normal. We note that Tz +B is not a Toeplitz operator.

Remark 5.9. In Theorem 5.1 we have seen that a 2-hyponormal Toeplitz completion of[Tz ?? Tz

]is

automatically normal. Consequently, from the viewpoint of k-hyponormality as a bridge betweenhyponormality and subnormality, there is no gap between the 2-hyponormality and the subnor-

mality of[Tz TϕTψ Tz

](ϕ,ψ ∈ H2). Of course there does exist a gap between the hyponormality and

the 2-hyponormality of[Tz TϕTψ Tz

]. To see this, let

Φ :=

[z z2 + 2z2

z2 + 2z2 z

].

Then Φ is normal and if we put K :=[

12z2

z2

12

], then Φ −K Φ∗ ∈ H2

M2and ||K||∞ = 1, so that TΦ

is hyponormal. But by Theorem 5.1, TΦ is not 2-hyponormal. However, we have not been able tocharacterize all hyponormal completions of

[Tz ?? Tz

]; this completion problem appears to be quite

difficult.

6. Open Problems

1. Nakazi-Takahashi’s Theorem for matrix-valued symbols. T. Nakazi and K. Takahashi [NT]have shown that if ϕ ∈ L∞ is such that Tϕ is a hyponormal operator whose self-commutator[T ∗ϕ, Tϕ] is of finite rank then there exists a finite Blaschke product b ∈ E(ϕ) such that

deg (b) = rank [T ∗ϕ, Tϕ].

What is the matrix-valued version of Nakazi and Takahashi’s Theorem ? A candidate is as follows: IfΦ ∈ L∞Mn

is such that TΦ is a hyponormal operator whose self-commutator [T ∗Φ, TΦ] is of finite rankthen there exists a finite Blaschke-Potapov product B ∈ E(Φ) such that deg (B) = rank [T ∗Φ, TΦ].We note that the degree of the finite Blaschke-Potapov product B is defined by

deg (B) := dimH(B) = deg (detB) , (6.1)


where the second equality follows from the well-known Fredholm theory of block Toeplitz operators[Do2] that

dimH(Θ) = dim kerTΘ∗ = −indexTΘ

= −indexTdet Θ= dim kerT

det Θ

= dim(H(det Θ)

)= deg

(det Θ

).

Thus we conjecture the following:

Conjecture 6.1. If Φ ∈ L∞Mnis such that TΦ is a hyponormal operator whose self-commutator

[T ∗Φ, TΦ] is of finite rank then there exists a finite Blaschke-Potapov product B ∈ E(Φ) such that

rank [T ∗Φ, TΦ] = deg(

detB)

.

On the other hand, in [NT], it was shown that if ϕ ∈ L∞ is such that Tϕ is subnormal andϕ = qϕ, where q is a finite Blaschke product then Tϕ is normal or analytic. We now we pose itsblock version:

Problem 6.2. If Φ ∈ L∞Mnis such that TΦ is subnormal and Φ = BΦ∗, where B is a a finite

Blaschke-Potapov product, does it follow that TΦ is normal or analytic ?

2. Subnormality of block Toeplitz operators. In Remark 4.7 we have shown that if the “coprime”condition of Theorem 4.5 is dropped, then Theorem 4.5 may fail. However we note that the examplegiven in Remark 4.7 is a direct sum of a normal Toeplitz operator and an analytic Toeplitz operator.Based on this observation, we have:

Problem 6.3. Let Φ ∈ L∞Mnbe a matrix-valued rational function. If TΦ and T 2

Φ are hyponormal,but TΦ is neither normal nor analytic, does it follow that TΦ is of the form

TΦ =

[TA 00 TB

](where TA is normal and TB is analytic) ?

It is well-known that if T ∈ B(H) is subnormal then ker [T ∗, T ] is invariant under T . Thus wemight be tempted to guess that if the condition “TΦ and T 2

Φ are hyponormal”is replaced by “TΦ ishyponormal and ker [T ∗Φ, TΦ] is invariant under TΦ,” then the answer to Problem 6.3 is affirmative.But this is not the case. Indeed, consider

TΦ =

[2U + U∗ U∗

U∗ 2U + U∗

].

Then a straightforward calculation shows that TΦ is hyponormal and ker [T ∗Φ, TΦ] is invariant underTΦ, but TΦ is never normal (cf. [CHL, Remark 3.9]). However, if the condition “TΦ and T 2

Φ arehyponormal” is strengthened to “TΦ is subnormal”, what conclusion do you draw ?

3. Subnormal completion problem. Theorem 5.1 provides the subnormal Toeplitz completion of[U∗ ?? U∗

](U is the shift on H2). (6.2)

Moreover Remark 5.8 shows that there is a normal non-Toeplitz completion of (6.2). However wewere unable to find all subnormal completions of (6.2).

Problem 6.4. Let U be the shift on H2. Complete the unspecified entries of the partial blockmatrix

[U∗ ?? U∗

]to make it subnormal.

On the other hand, Theorem 5.1 shows that the solution of the subnormal Toeplitz completionof[U∗ ?? U∗

]consists of Toeplitz operators with symbols which are both analytic or trigonometric

polynomials of degree 1. Hence we might expect that if the symbols of the specified Toeplitzoperators of (6.2) are co-analytic polynomials of degree two then the non-analytic solution of theunspecified entries consists of trigonometric polynomials of degree ≤ 2.


More generally, we have:

Problem 6.5. If Φ and Ψ are co-analytic polynomials of degree n, does it follow that the non-analyticsolution of the subnormal Toeplitz completion of the partial Toeplitz matrix

[TΦ ?? TΨ

]consists of

Toeplitz operators whose symbols are trigonometric polynomials of degree ≤ n ?

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Raul E. CurtoDepartment of Mathematics, University of Iowa, Iowa City, IA 52242, U.S.A.e-mail: [email protected]

In Sung HwangDepartment of Mathematics, Sungkyunkwan University, Suwon 440-746, Koreae-mail: [email protected]

Woo Young LeeDepartment of Mathematics, Seoul National University, Seoul 151-742, Koreae-mail: [email protected]

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